Find the value of $$\sin^2 60^\circ + \cos^2 30^\circ - \sin^2 45^\circ - 3 \sin^2 90^\circ$$.
$$\sin^260^{\circ}+\cos^230^{\circ}-\sin^245^{\circ}-3\sin^290^{\circ}=\left(\frac{\sqrt{3}}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2-\left(\frac{1}{\sqrt{2}}\right)^2-3\left(1\right)^2$$
$$=\frac{3}{4}+\frac{3}{4}-\frac{1}{2}-3$$
$$=\frac{3+3-2-12}{4}$$
$$=\frac{-8}{4}$$
$$=-2$$
Hence, the correct answer is Option D
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