Chapter 3: Trigonometric Functions

We use the conversion $$180^\circ = \pi$$ radians, so $$1^\circ = \dfrac{\pi}{180}$$ radians.

First express $$40^\circ\ 20'$$ entirely in degrees. Since $$1' = \left(\dfrac{1}{60}\right)^\circ$$, we have

$$40^\circ\ 20' = 40^\circ + \dfrac{20}{60}^\circ = 40^\circ + \dfrac{1}{3}^\circ = \dfrac{121}{3}^\circ$$

Convert to radians:

$$\dfrac{121}{3}^\circ = \dfrac{\pi}{180} \times \dfrac{121}{3} \text{ radian} = \dfrac{121\pi}{540} \text{ radian}$$

Hence $$40^\circ\ 20' = \dfrac{121\pi}{540}$$ radian.

Using $$\pi \text{ radian} = 180^\circ$$, we get $$1 \text{ radian} = \dfrac{180^\circ}{\pi}$$.

Hence

$$6 \text{ radian} = \dfrac{180}{\pi} \times 6 \text{ degree} = \dfrac{1080}{\pi} \text{ degree}$$

Taking $$\pi = \dfrac{22}{7}$$:

$$\dfrac{1080}{\pi} = 1080 \times \dfrac{7}{22} = \dfrac{7560}{22} = 343\dfrac{7}{11} \text{ degree}$$

Now $$\dfrac{7}{11}$$ degree $$= \dfrac{7}{11} \times 60' = \dfrac{420}{11}' = 38\dfrac{2}{11}'$$, and $$\dfrac{2}{11}' = \dfrac{2}{11} \times 60'' = \dfrac{120}{11}'' \approx 10.9''$$, i.e. about $$11''$$.

Therefore $$6 \text{ radian} \approx 343^\circ\ 38'\ 11''$$ (nearest second).

Let $$r$$ cm be the radius and $$\theta$$ the central angle (in radians) subtended by an arc of length $$\ell = 37.4$$ cm. The arc-length formula is

$$\theta = \dfrac{\ell}{r} \quad \text{i.e.} \quad \ell = r\,\theta$$

Convert $$60^\circ$$ to radians: $$\theta = 60 \times \dfrac{\pi}{180} = \dfrac{\pi}{3}$$ radian.

Therefore

$$r = \dfrac{\ell}{\theta} = \dfrac{37.4}{\pi/3} = \dfrac{37.4 \times 3}{\pi}$$

With $$\pi = \dfrac{22}{7}$$:

$$r = \dfrac{37.4 \times 3 \times 7}{22} = \dfrac{785.4}{22} = 35.7 \text{ cm}$$

In $$60$$ minutes the minute hand sweeps out one complete revolution, i.e. through an angle of $$360^\circ$$.

So in $$40$$ minutes it sweeps through

$$\theta = \dfrac{40}{60} \times 360^\circ = 240^\circ$$

Convert to radians:

$$\theta = 240 \times \dfrac{\pi}{180} = \dfrac{4\pi}{3} \text{ radian}$$

The tip of the minute hand traces an arc of length

$$\ell = r\,\theta = 1.5 \times \dfrac{4\pi}{3} = 2\pi \text{ cm}$$

Using $$\pi = 3.14$$:

$$\ell = 2 \times 3.14 = 6.28 \text{ cm}$$

Let the radii of the two circles be $$r_1$$ and $$r_2$$, and let each circle contain an arc of length $$\ell$$. Let the angles subtended at the centres be (in radians)

$$\theta_1 = 65 \times \dfrac{\pi}{180} = \dfrac{13\pi}{36}, \qquad \theta_2 = 110 \times \dfrac{\pi}{180} = \dfrac{22\pi}{36} = \dfrac{11\pi}{18}$$

Using $$\ell = r\,\theta$$, we have $$\ell = r_1\theta_1 = r_2\theta_2$$, so

$$\dfrac{r_1}{r_2} = \dfrac{\theta_2}{\theta_1} = \dfrac{11\pi/18}{13\pi/36} = \dfrac{11\pi}{18} \times \dfrac{36}{13\pi} = \dfrac{22}{13}$$

Therefore $$r_1 : r_2 = 22 : 13$$.

Number of revolutions per minute $$= 360$$.

Number of revolutions per second $$= \dfrac{360}{60} = 6$$.

In one revolution the wheel turns through $$2\pi$$ radians, so in $$6$$ revolutions it turns through

$$6 \times 2\pi = 12\pi \text{ radian}$$

Here $$r = 100$$ cm and arc length $$\ell = 22$$ cm. The angle in radians is

$$\theta = \dfrac{\ell}{r} = \dfrac{22}{100} = \dfrac{11}{50} \text{ radian}$$

Convert to degrees using $$1 \text{ radian} = \dfrac{180^\circ}{\pi}$$ with $$\pi = \dfrac{22}{7}$$:

$$\theta = \dfrac{11}{50} \times \dfrac{180}{\pi} \text{ degree} = \dfrac{11}{50} \times \dfrac{180 \times 7}{22} \text{ degree} = \dfrac{11 \times 180 \times 7}{50 \times 22} \text{ degree} = \dfrac{13860}{1100} \text{ degree} = 12.6^\circ$$

Now $$0.6^\circ = 0.6 \times 60' = 36'$$, so

$$\theta = 12^\circ\ 36'$$

The radius is $$r = \dfrac{40}{2} = 20$$ cm. Let $$AB$$ be the chord of length $$20$$ cm and let $$O$$ be the centre. In triangle $$OAB$$,

$$OA = OB = AB = 20 \text{ cm}$$

so $$\triangle OAB$$ is equilateral. Hence the central angle is

$$\angle AOB = 60^\circ = 60 \times \dfrac{\pi}{180} \text{ radian} = \dfrac{\pi}{3} \text{ radian}$$

The length of the minor arc $$\overset{\frown}{AB}$$ is therefore

$$\ell = r\,\theta = 20 \times \dfrac{\pi}{3} = \dfrac{20\pi}{3} \text{ cm}$$

Let the radii be $$r_1$$ and $$r_2$$, and let the equal arc length be $$\ell$$. Convert the angles to radians:

$$\theta_1 = 60^\circ = \dfrac{\pi}{3} \text{ radian}, \qquad \theta_2 = 75^\circ = \dfrac{5\pi}{12} \text{ radian}$$

Since $$\ell = r\theta$$ for each circle and the arc lengths are equal, $$r_1\theta_1 = r_2\theta_2$$, hence

$$\dfrac{r_1}{r_2} = \dfrac{\theta_2}{\theta_1} = \dfrac{5\pi/12}{\pi/3} = \dfrac{5\pi}{12} \times \dfrac{3}{\pi} = \dfrac{5}{4}$$

Therefore $$r_1 : r_2 = 5 : 4$$.

Given $$\cos x = -\dfrac{3}{5}$$ and $$x$$ in the third quadrant. In the third quadrant $$\sin x < 0$$, $$\cos x < 0$$, $$\tan x > 0$$.

From $$\sec x = \dfrac{1}{\cos x}$$:

$$\sec x = -\dfrac{5}{3}$$

From $$\sin^2 x + \cos^2 x = 1$$:

$$\sin^2 x = 1 - \dfrac{9}{25} = \dfrac{16}{25} \implies \sin x = \pm\dfrac{4}{5}$$

Since $$\sin x < 0$$ in the third quadrant, $$\sin x = -\dfrac{4}{5}$$.

Then

$$\operatorname{cosec} x = \dfrac{1}{\sin x} = -\dfrac{5}{4}$$

$$\tan x = \dfrac{\sin x}{\cos x} = \dfrac{-4/5}{-3/5} = \dfrac{4}{3}$$

$$\cot x = \dfrac{1}{\tan x} = \dfrac{3}{4}$$

Given $$\cot x = -\dfrac{5}{12}$$ and $$x$$ in the second quadrant. In Q2, $$\sin x > 0$$, $$\cos x < 0$$, $$\tan x < 0$$, $$\cot x < 0$$, $$\sec x < 0$$, $$\operatorname{cosec} x > 0$$.

From $$\tan x = \dfrac{1}{\cot x}$$:

$$\tan x = -\dfrac{12}{5}$$

From $$1 + \cot^2 x = \operatorname{cosec}^2 x$$:

$$\operatorname{cosec}^2 x = 1 + \dfrac{25}{144} = \dfrac{169}{144} \implies \operatorname{cosec} x = \pm\dfrac{13}{12}$$

In Q2, $$\operatorname{cosec} x > 0$$, so $$\operatorname{cosec} x = \dfrac{13}{12}$$, and

$$\sin x = \dfrac{1}{\operatorname{cosec} x} = \dfrac{12}{13}$$

From $$1 + \tan^2 x = \sec^2 x$$:

$$\sec^2 x = 1 + \dfrac{144}{25} = \dfrac{169}{25} \implies \sec x = \pm\dfrac{13}{5}$$

In Q2, $$\sec x < 0$$, so $$\sec x = -\dfrac{13}{5}$$, and

$$\cos x = -\dfrac{5}{13}$$

The sine function has period $$2\pi$$, so $$\sin(\theta + 2k\pi) = \sin\theta$$ for every integer $$k$$. Reduce $$\dfrac{31\pi}{3}$$ modulo $$2\pi$$:

$$\dfrac{31\pi}{3} = 10\pi + \dfrac{\pi}{3} = 5(2\pi) + \dfrac{\pi}{3}$$

Hence

$$\sin \dfrac{31\pi}{3} = \sin\left(5 \cdot 2\pi + \dfrac{\pi}{3}\right) = \sin \dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}$$

Cosine is an even function, so $$\cos(-\theta) = \cos\theta$$. Also $$\cos$$ has period $$360^\circ$$.

$$\cos(-1710^\circ) = \cos(1710^\circ)$$

Reduce modulo $$360^\circ$$: $$1710 = 4 \times 360 + 270 = 1440 + 270$$, so

$$\cos(1710^\circ) = \cos(4 \cdot 360^\circ + 270^\circ) = \cos 270^\circ = 0$$

Therefore $$\cos(-1710^\circ) = 0$$.

In Q3, $$\sin x < 0$$, $$\cos x < 0$$, $$\tan x > 0$$, $$\cot x > 0$$, $$\sec x < 0$$, $$\operatorname{cosec} x < 0$$.

$$\sec x = \dfrac{1}{\cos x} = -2$$

Using $$\sin^2 x + \cos^2 x = 1$$:

$$\sin^2 x = 1 - \dfrac{1}{4} = \dfrac{3}{4} \implies \sin x = \pm\dfrac{\sqrt{3}}{2}$$

Since $$\sin x < 0$$ in Q3, $$\sin x = -\dfrac{\sqrt{3}}{2}$$.

$$\operatorname{cosec} x = \dfrac{1}{\sin x} = -\dfrac{2}{\sqrt{3}}$$

$$\tan x = \dfrac{\sin x}{\cos x} = \dfrac{-\sqrt{3}/2}{-1/2} = \sqrt{3}$$

$$\cot x = \dfrac{1}{\tan x} = \dfrac{1}{\sqrt{3}}$$

In Q2, $$\sin x > 0$$, $$\cos x < 0$$, $$\tan x < 0$$, $$\cot x < 0$$, $$\sec x < 0$$, $$\operatorname{cosec} x > 0$$.

$$\operatorname{cosec} x = \dfrac{1}{\sin x} = \dfrac{5}{3}$$

Using $$\sin^2 x + \cos^2 x = 1$$:

$$\cos^2 x = 1 - \dfrac{9}{25} = \dfrac{16}{25} \implies \cos x = \pm\dfrac{4}{5}$$

In Q2, $$\cos x < 0$$, so $$\cos x = -\dfrac{4}{5}$$, and $$\sec x = -\dfrac{5}{4}$$.

$$\tan x = \dfrac{\sin x}{\cos x} = \dfrac{3/5}{-4/5} = -\dfrac{3}{4}$$

$$\cot x = -\dfrac{4}{3}$$

In Q3, $$\sin x < 0$$, $$\cos x < 0$$, $$\tan x > 0$$, $$\cot x > 0$$, $$\sec x < 0$$, $$\operatorname{cosec} x < 0$$.

$$\tan x = \dfrac{1}{\cot x} = \dfrac{4}{3}$$

Using $$1 + \cot^2 x = \operatorname{cosec}^2 x$$:

$$\operatorname{cosec}^2 x = 1 + \dfrac{9}{16} = \dfrac{25}{16} \implies \operatorname{cosec} x = \pm\dfrac{5}{4}$$

In Q3, $$\operatorname{cosec} x < 0$$, so $$\operatorname{cosec} x = -\dfrac{5}{4}$$, and $$\sin x = -\dfrac{4}{5}$$.

Using $$1 + \tan^2 x = \sec^2 x$$:

$$\sec^2 x = 1 + \dfrac{16}{9} = \dfrac{25}{9} \implies \sec x = \pm\dfrac{5}{3}$$

In Q3, $$\sec x < 0$$, so $$\sec x = -\dfrac{5}{3}$$, and $$\cos x = -\dfrac{3}{5}$$.

In Q4, $$\sin x < 0$$, $$\cos x > 0$$, $$\tan x < 0$$, $$\cot x < 0$$, $$\sec x > 0$$, $$\operatorname{cosec} x < 0$$.

$$\cos x = \dfrac{1}{\sec x} = \dfrac{5}{13}$$

Using $$\sin^2 x + \cos^2 x = 1$$:

$$\sin^2 x = 1 - \dfrac{25}{169} = \dfrac{144}{169} \implies \sin x = \pm\dfrac{12}{13}$$

In Q4, $$\sin x < 0$$, so $$\sin x = -\dfrac{12}{13}$$, and $$\operatorname{cosec} x = -\dfrac{13}{12}$$.

$$\tan x = \dfrac{\sin x}{\cos x} = \dfrac{-12/13}{5/13} = -\dfrac{12}{5}, \qquad \cot x = -\dfrac{5}{12}$$

In Q2, $$\sin x > 0$$, $$\cos x < 0$$, $$\tan x < 0$$, $$\cot x < 0$$, $$\sec x < 0$$, $$\operatorname{cosec} x > 0$$.

$$\cot x = \dfrac{1}{\tan x} = -\dfrac{12}{5}$$

Using $$1 + \tan^2 x = \sec^2 x$$:

$$\sec^2 x = 1 + \dfrac{25}{144} = \dfrac{169}{144} \implies \sec x = \pm\dfrac{13}{12}$$

In Q2, $$\sec x < 0$$, so $$\sec x = -\dfrac{13}{12}$$ and $$\cos x = -\dfrac{12}{13}$$.

$$\sin x = \tan x \cdot \cos x = \left(-\dfrac{5}{12}\right)\left(-\dfrac{12}{13}\right) = \dfrac{5}{13}$$

$$\operatorname{cosec} x = \dfrac{13}{5}$$

Sine has period $$360^\circ$$, so $$\sin(n \cdot 360^\circ + \theta) = \sin\theta$$ for every integer $$n$$.

$$765 = 2 \times 360 + 45$$, hence

$$\sin 765^\circ = \sin(2 \cdot 360^\circ + 45^\circ) = \sin 45^\circ = \dfrac{1}{\sqrt{2}}$$

Since $$\sin(-\theta) = -\sin\theta$$, we have $$\operatorname{cosec}(-\theta) = -\operatorname{cosec}\theta$$. Also $$\operatorname{cosec}$$ has period $$360^\circ$$.

$$\operatorname{cosec}(-1410^\circ) = -\operatorname{cosec}(1410^\circ)$$

Now $$1410 = 3 \times 360 + 330$$, so

$$\operatorname{cosec}(1410^\circ) = \operatorname{cosec}(330^\circ)$$

$$\sin 330^\circ = \sin(360^\circ - 30^\circ) = -\sin 30^\circ = -\dfrac{1}{2}$$

So $$\operatorname{cosec}(330^\circ) = -2$$, and therefore

$$\operatorname{cosec}(-1410^\circ) = -(-2) = 2$$

Tangent has period $$\pi$$ (and certainly period $$2\pi$$). Reduce $$\dfrac{19\pi}{3}$$ modulo $$2\pi$$:

$$\dfrac{19\pi}{3} = 6\pi + \dfrac{\pi}{3} = 3(2\pi) + \dfrac{\pi}{3}$$

Therefore

$$\tan \dfrac{19\pi}{3} = \tan\left(3 \cdot 2\pi + \dfrac{\pi}{3}\right) = \tan \dfrac{\pi}{3} = \sqrt{3}$$

Sine is odd: $$\sin(-\theta) = -\sin\theta$$. And $$\sin$$ has period $$2\pi$$.

$$\sin\left(-\dfrac{11\pi}{3}\right) = -\sin\dfrac{11\pi}{3}$$

Write $$\dfrac{11\pi}{3} = 4\pi - \dfrac{\pi}{3} = 2(2\pi) - \dfrac{\pi}{3}$$, so

$$\sin\dfrac{11\pi}{3} = \sin\left(2 \cdot 2\pi - \dfrac{\pi}{3}\right) = \sin\left(-\dfrac{\pi}{3}\right) = -\sin\dfrac{\pi}{3} = -\dfrac{\sqrt{3}}{2}$$

Therefore

$$\sin\left(-\dfrac{11\pi}{3}\right) = -\left(-\dfrac{\sqrt{3}}{2}\right) = \dfrac{\sqrt{3}}{2}$$

Cotangent is odd: $$\cot(-\theta) = -\cot\theta$$, and has period $$\pi$$ (so also $$2\pi$$).

$$\cot\left(-\dfrac{15\pi}{4}\right) = -\cot\dfrac{15\pi}{4}$$

Write $$\dfrac{15\pi}{4} = 4\pi - \dfrac{\pi}{4} = 2(2\pi) - \dfrac{\pi}{4}$$, so

$$\cot\dfrac{15\pi}{4} = \cot\left(2 \cdot 2\pi - \dfrac{\pi}{4}\right) = \cot\left(-\dfrac{\pi}{4}\right) = -\cot\dfrac{\pi}{4} = -1$$

Therefore

$$\cot\left(-\dfrac{15\pi}{4}\right) = -(-1) = 1$$

Evaluate each trigonometric value:

$$\sin \dfrac{\pi}{6} = \dfrac{1}{2}, \qquad \sec \dfrac{\pi}{3} = 2, \qquad \cot \dfrac{\pi}{4} = 1$$

For $$\sin \dfrac{5\pi}{6}$$ use $$\sin(\pi - \theta) = \sin\theta$$:

$$\sin \dfrac{5\pi}{6} = \sin\left(\pi - \dfrac{\pi}{6}\right) = \sin \dfrac{\pi}{6} = \dfrac{1}{2}$$

Substitute into the LHS:

$$\text{LHS} = 3 \cdot \dfrac{1}{2} \cdot 2 - 4 \cdot \dfrac{1}{2} \cdot 1 = 3 - 2 = 1 = \text{RHS}$$

Hence proved.

Write $$15^\circ = 45^\circ - 30^\circ$$ and use $$\sin(A-B) = \sin A \cos B - \cos A \sin B$$:

$$\sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ$$

$$= \dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{3}}{2} - \dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{2} = \dfrac{\sqrt{3} - 1}{2\sqrt{2}}$$

Tangent has period $$\pi$$, so

$$\tan \dfrac{13\pi}{12} = \tan\left(\pi + \dfrac{\pi}{12}\right) = \tan \dfrac{\pi}{12}$$

Now $$\dfrac{\pi}{12} = \dfrac{\pi}{4} - \dfrac{\pi}{6}$$. Using $$\tan(A-B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B}$$,

$$\tan \dfrac{\pi}{12} = \dfrac{\tan \dfrac{\pi}{4} - \tan \dfrac{\pi}{6}}{1 + \tan \dfrac{\pi}{4} \tan \dfrac{\pi}{6}} = \dfrac{1 - \dfrac{1}{\sqrt{3}}}{1 + \dfrac{1}{\sqrt{3}}} = \dfrac{\sqrt{3} - 1}{\sqrt{3} + 1}$$

Rationalise the denominator by multiplying by $$\dfrac{\sqrt{3} - 1}{\sqrt{3} - 1}$$:

$$= \dfrac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - 1^2} = \dfrac{3 - 2\sqrt{3} + 1}{2} = \dfrac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$$

Use the sum and difference formulas:

$$\sin(x+y) = \sin x \cos y + \cos x \sin y$$

$$\sin(x-y) = \sin x \cos y - \cos x \sin y$$

So

$$\dfrac{\sin(x+y)}{\sin(x-y)} = \dfrac{\sin x \cos y + \cos x \sin y}{\sin x \cos y - \cos x \sin y}$$

Divide numerator and denominator by $$\cos x \cos y$$ (assuming neither is zero):

$$= \dfrac{\dfrac{\sin x}{\cos x} + \dfrac{\sin y}{\cos y}}{\dfrac{\sin x}{\cos x} - \dfrac{\sin y}{\cos y}} = \dfrac{\tan x + \tan y}{\tan x - \tan y}$$

This is the RHS. Hence proved.

Write $$3x = 2x + x$$ and apply $$\tan(A+B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$$:

$$\tan 3x = \tan(2x + x) = \dfrac{\tan 2x + \tan x}{1 - \tan 2x \tan x}$$

Cross-multiply:

$$\tan 3x (1 - \tan 2x \tan x) = \tan 2x + \tan x$$

$$\tan 3x - \tan 3x \tan 2x \tan x = \tan 2x + \tan x$$

Rearrange:

$$\tan 3x - \tan 2x - \tan x = \tan 3x \tan 2x \tan x$$

i.e. $$\tan 3x \tan 2x \tan x = \tan 3x - \tan 2x - \tan x$$. Hence proved.

Use the sum-to-product identity $$\cos(A+B) + \cos(A-B) = 2 \cos A \cos B$$ with $$A = \dfrac{\pi}{4}$$ and $$B = x$$:

$$\cos\left(\dfrac{\pi}{4} + x\right) + \cos\left(\dfrac{\pi}{4} - x\right) = 2 \cos \dfrac{\pi}{4} \cos x$$

Since $$\cos \dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}}$$,

$$= 2 \cdot \dfrac{1}{\sqrt{2}} \cdot \cos x = \sqrt{2}\,\cos x = \text{RHS}$$

Hence proved.

Apply the sum-to-product identities

$$\cos C + \cos D = 2 \cos\dfrac{C+D}{2} \cos\dfrac{C-D}{2}, \qquad \sin C - \sin D = 2 \cos\dfrac{C+D}{2} \sin\dfrac{C-D}{2}$$

with $$C = 7x$$, $$D = 5x$$:

$$\cos 7x + \cos 5x = 2 \cos 6x \cos x$$

$$\sin 7x - \sin 5x = 2 \cos 6x \sin x$$

Therefore

$$\dfrac{\cos 7x + \cos 5x}{\sin 7x - \sin 5x} = \dfrac{2 \cos 6x \cos x}{2 \cos 6x \sin x} = \dfrac{\cos x}{\sin x} = \cot x$$

Hence proved.

Numerator. Group as $$(\sin 5x + \sin x) - 2\sin 3x$$ and use $$\sin C + \sin D = 2 \sin\dfrac{C+D}{2} \cos\dfrac{C-D}{2}$$:

$$\sin 5x + \sin x = 2 \sin 3x \cos 2x$$

Hence the numerator becomes

$$2 \sin 3x \cos 2x - 2 \sin 3x = 2 \sin 3x (\cos 2x - 1)$$

Using $$\cos 2x - 1 = -2 \sin^2 x$$,

$$\text{Numerator} = 2 \sin 3x \cdot (-2 \sin^2 x) = -4 \sin 3x \sin^2 x$$

Denominator. Use $$\cos C - \cos D = -2 \sin\dfrac{C+D}{2} \sin\dfrac{C-D}{2}$$:

$$\cos 5x - \cos x = -2 \sin 3x \sin 2x = -2 \sin 3x \cdot 2 \sin x \cos x = -4 \sin 3x \sin x \cos x$$

Ratio.

$$\dfrac{-4 \sin 3x \sin^2 x}{-4 \sin 3x \sin x \cos x} = \dfrac{\sin x}{\cos x} = \tan x$$

Hence proved.

Substitute the standard values

$$\sin \dfrac{\pi}{6} = \dfrac{1}{2}, \qquad \cos \dfrac{\pi}{3} = \dfrac{1}{2}, \qquad \tan \dfrac{\pi}{4} = 1$$

into the LHS:

$$\text{LHS} = \left(\dfrac{1}{2}\right)^2 + \left(\dfrac{1}{2}\right)^2 - (1)^2 = \dfrac{1}{4} + \dfrac{1}{4} - 1 = \dfrac{1}{2} - 1 = -\dfrac{1}{2} = \text{RHS}$$

Hence proved.

Compute each value.

$$\sin \dfrac{\pi}{6} = \dfrac{1}{2}, \qquad \cos \dfrac{\pi}{3} = \dfrac{1}{2}$$

For $$\operatorname{cosec}\dfrac{7\pi}{6}$$ use $$\sin\left(\pi + \dfrac{\pi}{6}\right) = -\sin\dfrac{\pi}{6} = -\dfrac{1}{2}$$, so

$$\operatorname{cosec}\dfrac{7\pi}{6} = \dfrac{1}{\sin(7\pi/6)} = -2, \quad \operatorname{cosec}^2 \dfrac{7\pi}{6} = 4$$

Now substitute:

$$\text{LHS} = 2 \cdot \dfrac{1}{4} + 4 \cdot \dfrac{1}{4} = \dfrac{1}{2} + 1 = \dfrac{3}{2} = \text{RHS}$$

Hence proved.

Standard values:

$$\cot \dfrac{\pi}{6} = \sqrt{3} \implies \cot^2 \dfrac{\pi}{6} = 3$$

$$\tan \dfrac{\pi}{6} = \dfrac{1}{\sqrt{3}} \implies \tan^2 \dfrac{\pi}{6} = \dfrac{1}{3}, \quad 3 \tan^2 \dfrac{\pi}{6} = 1$$

For $$\operatorname{cosec} \dfrac{5\pi}{6}$$ use $$\sin\left(\pi - \dfrac{\pi}{6}\right) = \sin \dfrac{\pi}{6} = \dfrac{1}{2}$$:

$$\operatorname{cosec} \dfrac{5\pi}{6} = 2$$

Therefore

$$\text{LHS} = 3 + 2 + 1 = 6 = \text{RHS}$$

Hence proved.

Compute each function value.

$$\sin \dfrac{3\pi}{4} = \sin\left(\pi - \dfrac{\pi}{4}\right) = \sin \dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}}, \quad \sin^2 \dfrac{3\pi}{4} = \dfrac{1}{2}$$

$$\cos \dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}}, \quad \cos^2 \dfrac{\pi}{4} = \dfrac{1}{2}$$

$$\sec \dfrac{\pi}{3} = 2, \quad \sec^2 \dfrac{\pi}{3} = 4$$

Substitute:

$$\text{LHS} = 2 \cdot \dfrac{1}{2} + 2 \cdot \dfrac{1}{2} + 2 \cdot 4 = 1 + 1 + 8 = 10 = \text{RHS}$$

Hence proved.

Use $$\cos(A+B) = \cos A \cos B - \sin A \sin B$$ with $$A = \dfrac{\pi}{4} - x$$ and $$B = \dfrac{\pi}{4} - y$$:

$$\text{LHS} = \cos\!\left[\left(\dfrac{\pi}{4} - x\right) + \left(\dfrac{\pi}{4} - y\right)\right] = \cos\!\left(\dfrac{\pi}{2} - (x+y)\right)$$

Since $$\cos\left(\dfrac{\pi}{2} - \theta\right) = \sin\theta$$,

$$\text{LHS} = \sin(x+y) = \text{RHS}$$

Hence proved.

Use $$\tan(A \pm B) = \dfrac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$$ with $$A = \dfrac{\pi}{4}$$, so $$\tan A = 1$$:

$$\tan\!\left(\dfrac{\pi}{4} + x\right) = \dfrac{1 + \tan x}{1 - \tan x}, \qquad \tan\!\left(\dfrac{\pi}{4} - x\right) = \dfrac{1 - \tan x}{1 + \tan x}$$

Divide:

$$\dfrac{\tan\!\left(\dfrac{\pi}{4} + x\right)}{\tan\!\left(\dfrac{\pi}{4} - x\right)} = \dfrac{1 + \tan x}{1 - \tan x} \times \dfrac{1 + \tan x}{1 - \tan x} = \left(\dfrac{1 + \tan x}{1 - \tan x}\right)^2 = \text{RHS}$$

Hence proved.

Apply the standard reductions:

$$\cos(\pi + x) = -\cos x, \quad \cos(-x) = \cos x$$

$$\sin(\pi - x) = \sin x, \quad \cos\!\left(\dfrac{\pi}{2} + x\right) = -\sin x$$

Substitute:

$$\text{LHS} = \dfrac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)} = \dfrac{-\cos^2 x}{-\sin^2 x} = \dfrac{\cos^2 x}{\sin^2 x} = \cot^2 x = \text{RHS}$$

Hence proved.

Reduce each factor using standard identities.

$$\cos\!\left(\dfrac{3\pi}{2} + x\right) = \cos\dfrac{3\pi}{2}\cos x - \sin\dfrac{3\pi}{2}\sin x = 0 \cdot \cos x - (-1)\sin x = \sin x$$

$$\cos(2\pi + x) = \cos x$$

$$\cot\!\left(\dfrac{3\pi}{2} - x\right) = \dfrac{\cos\!\left(\dfrac{3\pi}{2} - x\right)}{\sin\!\left(\dfrac{3\pi}{2} - x\right)} = \dfrac{-\sin x}{-\cos x} = \tan x$$

$$\cot(2\pi + x) = \cot x$$

So

$$\cot\!\left(\dfrac{3\pi}{2} - x\right) + \cot(2\pi + x) = \tan x + \cot x = \dfrac{\sin x}{\cos x} + \dfrac{\cos x}{\sin x} = \dfrac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \dfrac{1}{\sin x \cos x}$$

Multiply everything together:

$$\text{LHS} = \sin x \cdot \cos x \cdot \dfrac{1}{\sin x \cos x} = 1 = \text{RHS}$$

Hence proved.

Recognise the LHS as $$\cos A \cos B + \sin A \sin B = \cos(A - B)$$ with $$A = (n+2)x$$ and $$B = (n+1)x$$:

$$\text{LHS} = \cos\!\big((n+2)x - (n+1)x\big) = \cos x = \text{RHS}$$

Hence proved.

Use $$\cos C - \cos D = -2 \sin\dfrac{C+D}{2}\,\sin\dfrac{C-D}{2}$$ with $$C = \dfrac{3\pi}{4} + x$$, $$D = \dfrac{3\pi}{4} - x$$.

$$\dfrac{C+D}{2} = \dfrac{3\pi}{4}, \qquad \dfrac{C-D}{2} = x$$

So

$$\text{LHS} = -2 \sin \dfrac{3\pi}{4} \sin x = -2 \cdot \dfrac{1}{\sqrt{2}} \cdot \sin x = -\sqrt{2} \sin x = \text{RHS}$$

Hence proved.

Use the identity $$\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$$. (Derivation: $$\sin^2 A - \sin^2 B = (\sin A - \sin B)(\sin A + \sin B) = [2 \cos\frac{A+B}{2}\sin\frac{A-B}{2}][2 \sin\frac{A+B}{2}\cos\frac{A-B}{2}] = \sin(A+B)\sin(A-B)$$.)

With $$A = 6x$$, $$B = 4x$$:

$$\sin^2 6x - \sin^2 4x = \sin(6x + 4x)\sin(6x - 4x) = \sin 10x \sin 2x = \text{RHS}$$

Hence proved.

Use $$\cos^2 A - \cos^2 B = -\sin(A+B)\sin(A-B)$$. (This follows from $$\cos^2\theta = 1 - \sin^2\theta$$ and the identity in Q12.)

With $$A = 2x$$, $$B = 6x$$:

$$\cos^2 2x - \cos^2 6x = -\sin(2x + 6x)\sin(2x - 6x) = -\sin 8x \sin(-4x)$$

Since $$\sin(-4x) = -\sin 4x$$,

$$= -\sin 8x \cdot (-\sin 4x) = \sin 4x \sin 8x = \text{RHS}$$

Hence proved.

Group the first and third terms and use $$\sin C + \sin D = 2 \sin\dfrac{C+D}{2}\cos\dfrac{C-D}{2}$$:

$$\sin 2x + \sin 6x = 2 \sin 4x \cos 2x$$

So

$$\text{LHS} = 2 \sin 4x \cos 2x + 2 \sin 4x = 2 \sin 4x (\cos 2x + 1)$$

Using $$\cos 2x = 2 \cos^2 x - 1$$, we get $$\cos 2x + 1 = 2 \cos^2 x$$, so

$$\text{LHS} = 2 \sin 4x \cdot 2 \cos^2 x = 4 \cos^2 x \sin 4x = \text{RHS}$$

Hence proved.

Apply the sum-to-product identities:

$$\sin 5x + \sin 3x = 2 \sin 4x \cos x, \qquad \sin 5x - \sin 3x = 2 \cos 4x \sin x$$

LHS.

$$\cot 4x (\sin 5x + \sin 3x) = \dfrac{\cos 4x}{\sin 4x} \cdot 2 \sin 4x \cos x = 2 \cos 4x \cos x$$

RHS.

$$\cot x (\sin 5x - \sin 3x) = \dfrac{\cos x}{\sin x} \cdot 2 \cos 4x \sin x = 2 \cos x \cos 4x$$

So LHS $$= 2 \cos x \cos 4x =$$ RHS. Hence proved.

Use the difference-to-product identities:

$$\cos C - \cos D = -2 \sin\dfrac{C+D}{2}\sin\dfrac{C-D}{2}, \quad \sin C - \sin D = 2 \cos\dfrac{C+D}{2}\sin\dfrac{C-D}{2}$$

Numerator with $$C = 9x$$, $$D = 5x$$:

$$\cos 9x - \cos 5x = -2 \sin 7x \sin 2x$$

Denominator with $$C = 17x$$, $$D = 3x$$:

$$\sin 17x - \sin 3x = 2 \cos 10x \sin 7x$$

Therefore

$$\dfrac{\cos 9x - \cos 5x}{\sin 17x - \sin 3x} = \dfrac{-2 \sin 7x \sin 2x}{2 \cos 10x \sin 7x} = -\dfrac{\sin 2x}{\cos 10x} = \text{RHS}$$

Hence proved.

Apply the sum-to-product identities:

$$\sin 5x + \sin 3x = 2 \sin 4x \cos x$$

$$\cos 5x + \cos 3x = 2 \cos 4x \cos x$$

Divide:

$$\dfrac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \dfrac{2 \sin 4x \cos x}{2 \cos 4x \cos x} = \dfrac{\sin 4x}{\cos 4x} = \tan 4x = \text{RHS}$$

Hence proved.

Use the identities

$$\sin x - \sin y = 2 \cos\dfrac{x+y}{2}\sin\dfrac{x-y}{2}, \qquad \cos x + \cos y = 2 \cos\dfrac{x+y}{2}\cos\dfrac{x-y}{2}$$

Divide:

$$\dfrac{\sin x - \sin y}{\cos x + \cos y} = \dfrac{2 \cos\dfrac{x+y}{2}\sin\dfrac{x-y}{2}}{2 \cos\dfrac{x+y}{2}\cos\dfrac{x-y}{2}} = \dfrac{\sin\dfrac{x-y}{2}}{\cos\dfrac{x-y}{2}} = \tan\dfrac{x-y}{2} = \text{RHS}$$

Hence proved.

Sum-to-product:

$$\sin x + \sin 3x = 2 \sin 2x \cos x, \qquad \cos x + \cos 3x = 2 \cos 2x \cos x$$

Divide:

$$\dfrac{\sin x + \sin 3x}{\cos x + \cos 3x} = \dfrac{2 \sin 2x \cos x}{2 \cos 2x \cos x} = \tan 2x = \text{RHS}$$

Hence proved.

Numerator: use $$\sin C - \sin D = 2 \cos\dfrac{C+D}{2}\sin\dfrac{C-D}{2}$$ with $$C = x$$, $$D = 3x$$:

$$\sin x - \sin 3x = 2 \cos 2x \sin(-x) = -2 \cos 2x \sin x$$

Denominator: $$\sin^2 x - \cos^2 x = -(\cos^2 x - \sin^2 x) = -\cos 2x$$.

Ratio:

$$\dfrac{-2 \cos 2x \sin x}{-\cos 2x} = 2 \sin x = \text{RHS}$$

Hence proved.

Pair the outer terms in numerator and denominator.

Numerator:

$$\cos 4x + \cos 2x = 2 \cos 3x \cos x$$

so

$$\cos 4x + \cos 3x + \cos 2x = 2 \cos 3x \cos x + \cos 3x = \cos 3x (2 \cos x + 1)$$

Denominator:

$$\sin 4x + \sin 2x = 2 \sin 3x \cos x$$

so

$$\sin 4x + \sin 3x + \sin 2x = 2 \sin 3x \cos x + \sin 3x = \sin 3x (2 \cos x + 1)$$

Hence

$$\dfrac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \dfrac{\cos 3x (2\cos x + 1)}{\sin 3x (2 \cos x + 1)} = \dfrac{\cos 3x}{\sin 3x} = \cot 3x = \text{RHS}$$

Hence proved.

Write $$3x = 2x + x$$. Using $$\cot(A + B) = \dfrac{\cot A \cot B - 1}{\cot B + \cot A}$$ with $$A = 2x$$, $$B = x$$:

$$\cot 3x = \dfrac{\cot 2x \cot x - 1}{\cot x + \cot 2x}$$

Cross-multiply:

$$\cot 3x (\cot x + \cot 2x) = \cot 2x \cot x - 1$$

$$\cot 3x \cot x + \cot 3x \cot 2x = \cot 2x \cot x - 1$$

Rearrange:

$$\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1 = \text{RHS}$$

Hence proved.

Write $$4x = 2(2x)$$ and use $$\tan 2\theta = \dfrac{2 \tan\theta}{1 - \tan^2\theta}$$. Let $$t = \tan x$$.

Then

$$\tan 2x = \dfrac{2t}{1 - t^2}$$

and

$$\tan 4x = \dfrac{2 \tan 2x}{1 - \tan^2 2x}$$

Compute $$\tan^2 2x = \dfrac{4t^2}{(1 - t^2)^2}$$, so

$$1 - \tan^2 2x = \dfrac{(1 - t^2)^2 - 4t^2}{(1 - t^2)^2} = \dfrac{1 - 2t^2 + t^4 - 4t^2}{(1 - t^2)^2} = \dfrac{1 - 6t^2 + t^4}{(1 - t^2)^2}$$

Also $$2 \tan 2x = \dfrac{4t}{1 - t^2}$$. Therefore

$$\tan 4x = \dfrac{4t/(1 - t^2)}{(1 - 6t^2 + t^4)/(1 - t^2)^2} = \dfrac{4t}{1 - t^2} \cdot \dfrac{(1 - t^2)^2}{1 - 6t^2 + t^4} = \dfrac{4t (1 - t^2)}{1 - 6t^2 + t^4}$$

Restoring $$t = \tan x$$:

$$\tan 4x = \dfrac{4 \tan x (1 - \tan^2 x)}{1 - 6 \tan^2 x + \tan^4 x} = \text{RHS}$$

Hence proved.

Write $$4x = 2(2x)$$ and use $$\cos 2\theta = 1 - 2 \sin^2 \theta$$ with $$\theta = 2x$$:

$$\cos 4x = 1 - 2 \sin^2 2x$$

Substitute $$\sin 2x = 2 \sin x \cos x$$, so $$\sin^2 2x = 4 \sin^2 x \cos^2 x$$:

$$\cos 4x = 1 - 2 \cdot 4 \sin^2 x \cos^2 x = 1 - 8 \sin^2 x \cos^2 x = \text{RHS}$$

Hence proved.

Write $$6x = 3(2x)$$ and apply the triple-angle identity $$\cos 3\theta = 4 \cos^3 \theta - 3 \cos\theta$$ with $$\theta = 2x$$:

$$\cos 6x = 4 \cos^3 2x - 3 \cos 2x$$

Use $$\cos 2x = 2 \cos^2 x - 1$$. Let $$c = \cos^2 x$$, so $$\cos 2x = 2c - 1$$.

$$\cos^3 2x = (2c - 1)^3 = 8c^3 - 12c^2 + 6c - 1$$

$$4 \cos^3 2x = 32 c^3 - 48 c^2 + 24 c - 4$$

$$3 \cos 2x = 3(2c - 1) = 6c - 3$$

Therefore

$$\cos 6x = (32 c^3 - 48 c^2 + 24 c - 4) - (6c - 3) = 32 c^3 - 48 c^2 + 18 c - 1$$

Restoring $$c = \cos^2 x$$:

$$\cos 6x = 32 \cos^6 x - 48 \cos^4 x + 18 \cos^2 x - 1 = \text{RHS}$$

Hence proved.

First find $$\cos x$$ and $$\sin y$$. In Q2 we have $$\cos x < 0$$ and $$\sin y > 0$$.

From $$\sin^2 x + \cos^2 x = 1$$:

$$\cos^2 x = 1 - \dfrac{9}{25} = \dfrac{16}{25} \implies \cos x = -\dfrac{4}{5}$$

From $$\sin^2 y + \cos^2 y = 1$$:

$$\sin^2 y = 1 - \dfrac{144}{169} = \dfrac{25}{169} \implies \sin y = \dfrac{5}{13}$$

Using $$\sin(x+y) = \sin x \cos y + \cos x \sin y$$:

$$\sin(x+y) = \dfrac{3}{5} \cdot \left(-\dfrac{12}{13}\right) + \left(-\dfrac{4}{5}\right) \cdot \dfrac{5}{13}$$

$$= -\dfrac{36}{65} - \dfrac{20}{65} = -\dfrac{56}{65}$$

Use the product-to-sum formula $$\cos A \cos B = \dfrac{1}{2}\big[\cos(A-B) + \cos(A+B)\big]$$ on each term.

First term:

$$\cos 2x \cos\dfrac{x}{2} = \dfrac{1}{2}\left[\cos\dfrac{3x}{2} + \cos\dfrac{5x}{2}\right]$$

Second term:

$$\cos 3x \cos\dfrac{9x}{2} = \dfrac{1}{2}\left[\cos\!\left(3x - \dfrac{9x}{2}\right) + \cos\!\left(3x + \dfrac{9x}{2}\right)\right] = \dfrac{1}{2}\left[\cos\dfrac{3x}{2} + \cos\dfrac{15x}{2}\right]$$

(using $$\cos(-\theta) = \cos\theta$$ for the first piece.)

Subtract:

$$\text{LHS} = \dfrac{1}{2}\left[\cos\dfrac{3x}{2} + \cos\dfrac{5x}{2}\right] - \dfrac{1}{2}\left[\cos\dfrac{3x}{2} + \cos\dfrac{15x}{2}\right] = \dfrac{1}{2}\left[\cos\dfrac{5x}{2} - \cos\dfrac{15x}{2}\right]$$

Apply $$\cos C - \cos D = -2 \sin\dfrac{C+D}{2}\sin\dfrac{C-D}{2}$$ with $$C = \dfrac{5x}{2}$$, $$D = \dfrac{15x}{2}$$:

$$\dfrac{C+D}{2} = 5x, \qquad \dfrac{C-D}{2} = -\dfrac{5x}{2}$$

$$\cos\dfrac{5x}{2} - \cos\dfrac{15x}{2} = -2 \sin 5x \sin\!\left(-\dfrac{5x}{2}\right) = 2 \sin 5x \sin\dfrac{5x}{2}$$

Therefore

$$\text{LHS} = \dfrac{1}{2} \cdot 2 \sin 5x \sin\dfrac{5x}{2} = \sin 5x \sin\dfrac{5x}{2} = \text{RHS}$$

Hence proved.

Let $$\theta = \dfrac{\pi}{8}$$, so $$2\theta = \dfrac{\pi}{4}$$ and $$\tan 2\theta = 1$$. Using $$\tan 2\theta = \dfrac{2 \tan\theta}{1 - \tan^2 \theta}$$:

$$\dfrac{2 \tan\theta}{1 - \tan^2\theta} = 1$$

$$2 \tan\theta = 1 - \tan^2\theta$$

$$\tan^2\theta + 2 \tan\theta - 1 = 0$$

By the quadratic formula,

$$\tan\theta = \dfrac{-2 \pm \sqrt{4 + 4}}{2} = \dfrac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2}$$

Since $$0 < \dfrac{\pi}{8} < \dfrac{\pi}{2}$$, we have $$\tan\theta > 0$$. Hence

$$\tan\dfrac{\pi}{8} = \sqrt{2} - 1$$

Since $$\pi < x < \dfrac{3\pi}{2}$$, $$x$$ is in Q3, so $$\sin x < 0$$ and $$\cos x < 0$$. Dividing by $$2$$ gives $$\dfrac{\pi}{2} < \dfrac{x}{2} < \dfrac{3\pi}{4}$$, so $$\dfrac{x}{2}$$ is in Q2, with $$\sin\dfrac{x}{2} > 0$$ and $$\cos\dfrac{x}{2} < 0$$.

Find $$\cos x$$ and $$\sin x$$. From $$\sec^2 x = 1 + \tan^2 x = 1 + \dfrac{9}{16} = \dfrac{25}{16}$$, $$\sec x = \pm\dfrac{5}{4}$$. In Q3, $$\sec x < 0$$:

$$\sec x = -\dfrac{5}{4} \implies \cos x = -\dfrac{4}{5}$$

$$\sin x = \tan x \cdot \cos x = \dfrac{3}{4} \cdot \left(-\dfrac{4}{5}\right) = -\dfrac{3}{5}$$

Now use the half-angle formulas:

$$\sin^2 \dfrac{x}{2} = \dfrac{1 - \cos x}{2} = \dfrac{1 - (-4/5)}{2} = \dfrac{9/5}{2} = \dfrac{9}{10} \implies \sin\dfrac{x}{2} = \dfrac{3}{\sqrt{10}}$$

(positive since $$x/2$$ is in Q2).

$$\cos^2 \dfrac{x}{2} = \dfrac{1 + \cos x}{2} = \dfrac{1 - 4/5}{2} = \dfrac{1/5}{2} = \dfrac{1}{10} \implies \cos\dfrac{x}{2} = -\dfrac{1}{\sqrt{10}}$$

(negative in Q2).

$$\tan\dfrac{x}{2} = \dfrac{\sin(x/2)}{\cos(x/2)} = \dfrac{3/\sqrt{10}}{-1/\sqrt{10}} = -3$$

Use $$\cos^2 \theta = \dfrac{1 + \cos 2\theta}{2}$$:

$$\text{LHS} = \dfrac{1 + \cos 2x}{2} + \dfrac{1 + \cos\!\left(2x + \dfrac{2\pi}{3}\right)}{2} + \dfrac{1 + \cos\!\left(2x - \dfrac{2\pi}{3}\right)}{2}$$

$$= \dfrac{3}{2} + \dfrac{1}{2}\Big[\cos 2x + \cos\!\left(2x + \dfrac{2\pi}{3}\right) + \cos\!\left(2x - \dfrac{2\pi}{3}\right)\Big]$$

Apply $$\cos(A+B) + \cos(A-B) = 2 \cos A \cos B$$ to the last two terms with $$A = 2x$$, $$B = \dfrac{2\pi}{3}$$:

$$\cos\!\left(2x + \dfrac{2\pi}{3}\right) + \cos\!\left(2x - \dfrac{2\pi}{3}\right) = 2 \cos 2x \cos\dfrac{2\pi}{3} = 2 \cos 2x \cdot \left(-\dfrac{1}{2}\right) = -\cos 2x$$

The bracket therefore equals $$\cos 2x - \cos 2x = 0$$, and

$$\text{LHS} = \dfrac{3}{2} + 0 = \dfrac{3}{2} = \text{RHS}$$

Hence proved.

Apply $$2 \cos A \cos B = \cos(A+B) + \cos(A-B)$$ with $$A = \dfrac{\pi}{13}$$, $$B = \dfrac{9\pi}{13}$$:

$$2 \cos\dfrac{\pi}{13} \cos\dfrac{9\pi}{13} = \cos\dfrac{10\pi}{13} + \cos\!\left(-\dfrac{8\pi}{13}\right) = \cos\dfrac{10\pi}{13} + \cos\dfrac{8\pi}{13}$$

So

$$\text{LHS} = \cos\dfrac{10\pi}{13} + \cos\dfrac{8\pi}{13} + \cos\dfrac{3\pi}{13} + \cos\dfrac{5\pi}{13}$$

Now use $$\cos(\pi - \theta) = -\cos\theta$$:

$$\cos\dfrac{10\pi}{13} = \cos\!\left(\pi - \dfrac{3\pi}{13}\right) = -\cos\dfrac{3\pi}{13}$$

$$\cos\dfrac{8\pi}{13} = \cos\!\left(\pi - \dfrac{5\pi}{13}\right) = -\cos\dfrac{5\pi}{13}$$

Therefore

$$\text{LHS} = -\cos\dfrac{3\pi}{13} - \cos\dfrac{5\pi}{13} + \cos\dfrac{3\pi}{13} + \cos\dfrac{5\pi}{13} = 0 = \text{RHS}$$

Hence proved.

Expand and regroup:

$$\text{LHS} = \sin 3x \sin x + \sin^2 x + \cos 3x \cos x - \cos^2 x$$

$$= (\sin 3x \sin x + \cos 3x \cos x) - (\cos^2 x - \sin^2 x)$$

The first bracket is $$\cos(3x - x) = \cos 2x$$ (using $$\cos A \cos B + \sin A \sin B = \cos(A - B)$$), and the second bracket equals $$\cos 2x$$. Therefore

$$\text{LHS} = \cos 2x - \cos 2x = 0 = \text{RHS}$$

Hence proved.

Expand the squares:

$$(\cos x + \cos y)^2 = \cos^2 x + 2 \cos x \cos y + \cos^2 y$$

$$(\sin x - \sin y)^2 = \sin^2 x - 2 \sin x \sin y + \sin^2 y$$

Add and use $$\sin^2 + \cos^2 = 1$$:

$$\text{LHS} = (\cos^2 x + \sin^2 x) + (\cos^2 y + \sin^2 y) + 2(\cos x \cos y - \sin x \sin y)$$

$$= 1 + 1 + 2 \cos(x + y) = 2\big(1 + \cos(x + y)\big)$$

Using $$1 + \cos\theta = 2 \cos^2\dfrac{\theta}{2}$$ with $$\theta = x + y$$:

$$\text{LHS} = 2 \cdot 2 \cos^2\dfrac{x+y}{2} = 4 \cos^2\dfrac{x+y}{2} = \text{RHS}$$

Hence proved.

Expand the squares:

$$(\cos x - \cos y)^2 = \cos^2 x - 2 \cos x \cos y + \cos^2 y$$

$$(\sin x - \sin y)^2 = \sin^2 x - 2 \sin x \sin y + \sin^2 y$$

Add:

$$\text{LHS} = (\cos^2 x + \sin^2 x) + (\cos^2 y + \sin^2 y) - 2(\cos x \cos y + \sin x \sin y)$$

$$= 1 + 1 - 2 \cos(x - y) = 2\big(1 - \cos(x - y)\big)$$

Using $$1 - \cos\theta = 2 \sin^2\dfrac{\theta}{2}$$ with $$\theta = x - y$$:

$$\text{LHS} = 2 \cdot 2 \sin^2\dfrac{x-y}{2} = 4 \sin^2\dfrac{x-y}{2} = \text{RHS}$$

Hence proved.

Pair the outermost and the inner two terms and use $$\sin C + \sin D = 2 \sin\dfrac{C+D}{2}\cos\dfrac{C-D}{2}$$.

$$\sin x + \sin 7x = 2 \sin 4x \cos 3x$$

$$\sin 3x + \sin 5x = 2 \sin 4x \cos x$$

Add:

$$\text{LHS} = 2 \sin 4x (\cos 3x + \cos x)$$

Now $$\cos 3x + \cos x = 2 \cos 2x \cos x$$, so

$$\text{LHS} = 2 \sin 4x \cdot 2 \cos 2x \cos x = 4 \cos x \cos 2x \sin 4x = \text{RHS}$$

Hence proved.

Use the sum-to-product identities.

Numerator:

$$\sin 7x + \sin 5x = 2 \sin 6x \cos x, \qquad \sin 9x + \sin 3x = 2 \sin 6x \cos 3x$$

So the numerator equals

$$2 \sin 6x (\cos x + \cos 3x) = 2 \sin 6x \cdot 2 \cos 2x \cos x = 4 \sin 6x \cos 2x \cos x$$

Denominator:

$$\cos 7x + \cos 5x = 2 \cos 6x \cos x, \qquad \cos 9x + \cos 3x = 2 \cos 6x \cos 3x$$

So the denominator equals

$$2 \cos 6x (\cos x + \cos 3x) = 2 \cos 6x \cdot 2 \cos 2x \cos x = 4 \cos 6x \cos 2x \cos x$$

Divide:

$$\dfrac{4 \sin 6x \cos 2x \cos x}{4 \cos 6x \cos 2x \cos x} = \dfrac{\sin 6x}{\cos 6x} = \tan 6x = \text{RHS}$$

Hence proved.

Apply $$\sin C - \sin D = 2 \cos\dfrac{C+D}{2}\sin\dfrac{C-D}{2}$$ to $$\sin 3x - \sin x$$:

$$\sin 3x - \sin x = 2 \cos 2x \sin x$$

Also $$\sin 2x = 2 \sin x \cos x$$. Hence

$$\text{LHS} = (\sin 3x - \sin x) + \sin 2x = 2 \cos 2x \sin x + 2 \sin x \cos x = 2 \sin x (\cos 2x + \cos x)$$

Apply $$\cos A + \cos B = 2 \cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2}$$ to $$\cos 2x + \cos x$$:

$$\cos 2x + \cos x = 2 \cos\dfrac{3x}{2}\cos\dfrac{x}{2}$$

Therefore

$$\text{LHS} = 2 \sin x \cdot 2 \cos\dfrac{3x}{2}\cos\dfrac{x}{2} = 4 \sin x \cos\dfrac{x}{2}\cos\dfrac{3x}{2} = \text{RHS}$$

Hence proved.

$$x$$ in Q2 means $$\dfrac{\pi}{2} < x < \pi$$, so $$\dfrac{\pi}{4} < \dfrac{x}{2} < \dfrac{\pi}{2}$$ and $$\dfrac{x}{2}$$ is in Q1. Hence $$\sin\dfrac{x}{2}, \cos\dfrac{x}{2}, \tan\dfrac{x}{2}$$ are all positive.

Find $$\cos x$$ from $$\sec^2 x = 1 + \tan^2 x = 1 + \dfrac{16}{9} = \dfrac{25}{9}$$, so $$\sec x = \pm\dfrac{5}{3}$$. In Q2, $$\sec x < 0$$:

$$\sec x = -\dfrac{5}{3} \implies \cos x = -\dfrac{3}{5}$$

Use the half-angle formulas:

$$\sin^2\dfrac{x}{2} = \dfrac{1 - \cos x}{2} = \dfrac{1 + 3/5}{2} = \dfrac{8/5}{2} = \dfrac{4}{5} \implies \sin\dfrac{x}{2} = \dfrac{2}{\sqrt{5}}$$

$$\cos^2\dfrac{x}{2} = \dfrac{1 + \cos x}{2} = \dfrac{1 - 3/5}{2} = \dfrac{2/5}{2} = \dfrac{1}{5} \implies \cos\dfrac{x}{2} = \dfrac{1}{\sqrt{5}}$$

$$\tan\dfrac{x}{2} = \dfrac{\sin(x/2)}{\cos(x/2)} = \dfrac{2/\sqrt{5}}{1/\sqrt{5}} = 2$$

$$x$$ in Q3 means $$\pi < x < \dfrac{3\pi}{2}$$, so $$\dfrac{\pi}{2} < \dfrac{x}{2} < \dfrac{3\pi}{4}$$ and $$\dfrac{x}{2}$$ is in Q2. Hence $$\sin\dfrac{x}{2} > 0$$, $$\cos\dfrac{x}{2} < 0$$, $$\tan\dfrac{x}{2} < 0$$.

Use the half-angle formulas with $$\cos x = -\dfrac{1}{3}$$.

$$\sin^2\dfrac{x}{2} = \dfrac{1 - \cos x}{2} = \dfrac{1 + 1/3}{2} = \dfrac{4/3}{2} = \dfrac{2}{3} \implies \sin\dfrac{x}{2} = \sqrt{\dfrac{2}{3}} = \dfrac{\sqrt{6}}{3}$$

$$\cos^2\dfrac{x}{2} = \dfrac{1 + \cos x}{2} = \dfrac{1 - 1/3}{2} = \dfrac{2/3}{2} = \dfrac{1}{3} \implies \cos\dfrac{x}{2} = -\sqrt{\dfrac{1}{3}} = -\dfrac{1}{\sqrt{3}}$$

$$\tan\dfrac{x}{2} = \dfrac{\sin(x/2)}{\cos(x/2)} = \dfrac{\sqrt{2/3}}{-\sqrt{1/3}} = -\sqrt{\dfrac{2/3}{1/3}} = -\sqrt{2}$$

$$x$$ in Q2 means $$\dfrac{\pi}{2} < x < \pi$$, so $$\dfrac{\pi}{4} < \dfrac{x}{2} < \dfrac{\pi}{2}$$ and $$\dfrac{x}{2}$$ is in Q1. All three of $$\sin\dfrac{x}{2}, \cos\dfrac{x}{2}, \tan\dfrac{x}{2}$$ are positive.

Find $$\cos x$$: in Q2, $$\cos x < 0$$, and

$$\cos^2 x = 1 - \sin^2 x = 1 - \dfrac{1}{16} = \dfrac{15}{16} \implies \cos x = -\dfrac{\sqrt{15}}{4}$$

Half-angle formulas:

$$\sin^2\dfrac{x}{2} = \dfrac{1 - \cos x}{2} = \dfrac{1 + \dfrac{\sqrt{15}}{4}}{2} = \dfrac{4 + \sqrt{15}}{8}$$

$$\cos^2\dfrac{x}{2} = \dfrac{1 + \cos x}{2} = \dfrac{1 - \dfrac{\sqrt{15}}{4}}{2} = \dfrac{4 - \sqrt{15}}{8}$$

Simplify the surds. Note that

$$8 + 2\sqrt{15} = 5 + 3 + 2\sqrt{15} = (\sqrt{5} + \sqrt{3})^2$$

$$8 - 2\sqrt{15} = 5 + 3 - 2\sqrt{15} = (\sqrt{5} - \sqrt{3})^2$$

So

$$\sin\dfrac{x}{2} = \sqrt{\dfrac{4 + \sqrt{15}}{8}} = \sqrt{\dfrac{8 + 2\sqrt{15}}{16}} = \dfrac{\sqrt{5} + \sqrt{3}}{4}$$

$$\cos\dfrac{x}{2} = \sqrt{\dfrac{4 - \sqrt{15}}{8}} = \sqrt{\dfrac{8 - 2\sqrt{15}}{16}} = \dfrac{\sqrt{5} - \sqrt{3}}{4}$$

$$\tan\dfrac{x}{2} = \dfrac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} = \dfrac{(\sqrt{5} + \sqrt{3})^2}{(\sqrt{5})^2 - (\sqrt{3})^2} = \dfrac{5 + 2\sqrt{15} + 3}{5 - 3} = \dfrac{8 + 2\sqrt{15}}{2} = 4 + \sqrt{15}$$