Sum-to-Product Identities

We know that $$\cos\theta=\sin(90-\theta)$$

Thus, $$\cos 48^\circ=\sin (90-48)^\circ=\sin 42^\circ$$

$$\dfrac{\sin^{2}42^{o}-\sin^{2}12^{o}}{\sin^{2}24^{o}-\sin^{2}6^{o}}$$

$$\dfrac{\left(\sin42^o-\sin12^o\right)\left(\sin42^o+\sin12^o\right)}{\left(\sin24^o-\sin6^o\right)\left(\sin24^o+\sin6^o\right)}$$

Now, we know that - 

$$\sin A-\sin B=2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$$

$$\sin A+\sin B=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$$

Thus, the given equation will become - 

$$\dfrac{2\cos27^o\sin15^o\times2\sin27^o\cos15^o}{2\cos15^o\sin9^o\times2\sin15^o\cos9^o}$$  (Also, $$2\sin A\cos A = \sin 2A$$) 

$$\dfrac{\sin54^o\times\sin30^o}{\sin30^o\times\sin18^o}$$

$$\dfrac{\sin54^o}{\sin18^o}$$

Also, $$\sin54^o=\dfrac{\sqrt{5}+1}{4}$$ and $$\sin18^o=\dfrac{\sqrt{5}-1}{4}$$. Therefore, the fraction becomes - 

$$\dfrac{\frac{\sqrt{5}+1}{4}}{\frac{\sqrt{5}-1}{4}}=\dfrac{\sqrt{5}+1}{\sqrt{5}-1}\times\dfrac{\sqrt{5}+1}{\sqrt{5}+1}=\dfrac{\left(\sqrt{5}+1\right)^2}{4}=\dfrac{6+2\sqrt{5}}{4}=\dfrac{3+\sqrt{5}}{2}$$

Thus, $$\alpha=3$$ and $$\beta=1$$

$$\alpha+\beta=3+1=4$$

Step 1: Grouping and Converting to Sine and Cosine

We start by bringing the single tangent term to the denominator on the Left Hand Side to group our variables logically.

$$\frac{\tan(x + 100^\circ)}{\tan x} = \tan(x + 50^\circ) \tan(x - 50^\circ)$$

Next, we break everything down into sines and cosines using the fundamental identity $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$.

For the Left Hand Side, we have a fraction divided by a fraction. When dividing by a fraction, we multiply by its reciprocal.

$$\text{Left Side} = \frac{\frac{\sin(x+100^\circ)}{\cos(x+100^\circ)}}{\frac{\sin x}{\cos x}}$$
$$\text{Left Side} = \frac{\sin(x+100^\circ)}{\cos(x+100^\circ)} \times \frac{\cos x}{\sin x} = \frac{\sin(x+100^\circ) \cos x}{\cos(x+100^\circ) \sin x}$$

For the Right Hand Side, we simply multiply two tangent fractions straight across the numerators and denominators.

$$\text{Right Side} = \left( \frac{\sin(x+50^\circ)}{\cos(x+50^\circ)} \right) \left( \frac{\sin(x-50^\circ)}{\cos(x-50^\circ)} \right) = \frac{\sin(x+50^\circ) \sin(x-50^\circ)}{\cos(x+50^\circ) \cos(x-50^\circ)}$$

Equating our newly expanded sides gives us our working foundation:

$$\frac{\sin(x+100^\circ) \cos x}{\cos(x+100^\circ) \sin x} = \frac{\sin(x+50^\circ) \sin(x-50^\circ)}{\cos(x+50^\circ) \cos(x-50^\circ)}$$

Step 2: Applying Componendo and Dividendo

If $$\frac{a}{b} = \frac{c}{d}$$, we can instantly rewrite it as $$\frac{a+b}{a-b} = \frac{c+d}{c-d}$$. This allows us to rapidly build standard trigonometric identities.

Processing the Left Hand Side:

The new numerator becomes $$a+b$$:

$$\sin(x+100^\circ) \cos x + \cos(x+100^\circ) \sin x = \sin(x+100^\circ + x) = \sin(2x + 100^\circ)$$

The new denominator becomes $$a-b$$:

$$\sin(x+100^\circ) \cos x - \cos(x+100^\circ) \sin x = \sin(x+100^\circ - x) = \sin 100^\circ$$

Processing the Right Hand Side:

The new numerator becomes $$c+d$$:

$$\sin(x+50^\circ) \sin(x-50^\circ) + \cos(x+50^\circ) \cos(x-50^\circ) = \cos((x+50^\circ) - (x-50^\circ)) = \cos 100^\circ$$

The new denominator becomes $$c-d$$:

$$\sin(x+50^\circ) \sin(x-50^\circ) - \cos(x+50^\circ) \cos(x-50^\circ)$$
$$= -[ \cos(x+50^\circ) \cos(x-50^\circ) - \sin(x+50^\circ) \sin(x-50^\circ) ] = -\cos(x+50^\circ + x-50^\circ) = -\cos 2x$$

Putting the new simplified fractions together, our massive equation shrinks down to:

$$\frac{\sin(2x + 100^\circ)}{\sin 100^\circ} = \frac{\cos 100^\circ}{-\cos 2x}$$

Step 3: Cross Multiplication and Double Angle Identities

Cross multiply to remove the fractions completely:

$$-\sin(2x + 100^\circ) \cos 2x = \sin 100^\circ \cos 100^\circ$$

Multiply both sides of the equation by 2. This perfectly sets up the product to sum formula $$2\sin A \cos B = \sin(A+B) + \sin(A-B)$$ on the left, and the double angle formula $$2\sin \theta \cos \theta = \sin 2\theta$$ on the right.

$$-2\sin(2x + 100^\circ) \cos 2x = 2\sin 100^\circ \cos 100^\circ$$
$$-[ \sin(2x + 100^\circ + 2x) + \sin(2x + 100^\circ - 2x) ] = \sin 200^\circ$$
$$-[ \sin(4x + 100^\circ) + \sin 100^\circ ] = \sin 200^\circ$$

Move $$\sin 100^\circ$$ to the other side to isolate our $$x$$ variable:

$$-\sin(4x + 100^\circ) = \sin 200^\circ + \sin 100^\circ$$

Step 4: Sum to Product Conversion

We simplify the right side using the formula $$\sin C + \sin D = 2\sin\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)$$:

$$\sin 200^\circ + \sin 100^\circ = 2\sin\left(\frac{300^\circ}{2}\right)\cos\left(\frac{100^\circ}{2}\right) = 2\sin 150^\circ \cos 50^\circ$$

Because $$\sin 150^\circ = \frac{1}{2}$$, this entire term collapses cleanly into just $$\cos 50^\circ$$.

$$-\sin(4x + 100^\circ) = \cos 50^\circ$$
$$\sin(4x + 100^\circ) = -\cos 50^\circ$$

Step 5: Solving for the Final Angles

To solve the equation, we need matching trigonometric functions. We must convert the negative cosine into a sine function.

$$-\cos 50^\circ = -\sin 40^\circ = \sin(-40^\circ)$$

This gives us our final trigonometric equation to solve:

$$\sin(4x + 100^\circ) = \sin(-40^\circ)$$

We plug this into the standard general solution formula $$\theta = n(180^\circ) + (-1)^n \alpha$$:

$$4x + 100^\circ = n(180^\circ) + (-1)^n (-40^\circ)$$

Now we test consecutive integer values for $$n$$ to discover the specific solutions that fit into the given interval between $$0^\circ$$ and $$180^\circ$$.

• For n = 1:

  $$4x + 100^\circ = 180^\circ - (-40^\circ) = 220^\circ$$
  $$4x = 120^\circ \implies x = 30^\circ$$

• For n = 2:

  $$4x + 100^\circ = 360^\circ + (-40^\circ) = 320^\circ$$
  $$4x = 220^\circ \implies x = 55^\circ$$

• For n = 3:

  $$4x + 100^\circ = 540^\circ - (-40^\circ) = 580^\circ$$
  $$4x = 480^\circ \implies x = 120^\circ$$

• For n = 4:

  $$4x + 100^\circ = 720^\circ + (-40^\circ) = 680^\circ$$
  $$4x = 580^\circ \implies x = 145^\circ$$

If we try $$n = 0$$, we get $$x = -35^\circ$$, and if we try $$n = 5$$, we get $$x = 210^\circ$$. Both of these fall strictly outside the required domain boundaries.

The valid angles are exactly $$30^\circ$$, $$55^\circ$$, $$120^\circ$$, and $$145^\circ$$.

The final number of elements in the set is 4.

We are given $$\cos x + \cos y - \cos(x + y) = \frac{3}{2}$$ with $$0 < x, y < \pi$$.

Using the identities $$\cos x + \cos y = 2\cos\frac{x+y}{2}\cos\frac{x-y}{2}$$ and $$\cos(x+y) = 2\cos^2\frac{x+y}{2} - 1$$, we substitute to get $$2\cos\frac{x+y}{2}\cos\frac{x-y}{2} - 2\cos^2\frac{x+y}{2} + 1 = \frac{3}{2}$$.

Let $$u = \cos\frac{x+y}{2}$$ and $$v = \cos\frac{x-y}{2}$$. Then $$2uv - 2u^2 = \frac{1}{2}$$, or equivalently $$4u^2 - 4uv + 1 = 0$$.

Treating this as a quadratic in $$u$$: $$4u^2 - 4vu + 1 = 0$$. The discriminant must be non-negative: $$16v^2 - 16 \geq 0$$, so $$v^2 \geq 1$$. Since $$v = \cos\frac{x-y}{2}$$ and $$|v| \leq 1$$, we need $$v = \pm 1$$. Given the constraint $$0 < x, y < \pi$$, we have $$\left|\frac{x-y}{2}\right| < \frac{\pi}{2}$$, so $$v > 0$$, giving $$v = 1$$, which means $$x = y$$.

With $$x = y$$, the equation becomes $$2\cos x - \cos 2x = \frac{3}{2}$$, i.e., $$2\cos x - (2\cos^2 x - 1) = \frac{3}{2}$$, which simplifies to $$2\cos^2 x - 2\cos x + \frac{1}{2} = 0$$, or $$4\cos^2 x - 4\cos x + 1 = 0$$.

This gives $$(2\cos x - 1)^2 = 0$$, so $$\cos x = \frac{1}{2}$$, meaning $$x = \frac{\pi}{3}$$. Therefore $$x = y = \frac{\pi}{3}$$.

Thus, $$\sin x + \cos y = \sin\frac{\pi}{3} + \cos\frac{\pi}{3} = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{1 + \sqrt{3}}{2}$$.