MCQ
The value of $\cos^248^\circ-\sin^212^\circ$ is:
- A$\frac{\sqrt{5}+1}{8}$
- B$\frac{\sqrt{5}-1}{8}$
- C$\frac{\sqrt{5}+1}{5}$
- D$\frac{\sqrt{5}+1}{2\sqrt{2}}$
Solution:
$\cos^248^\circ-\sin^212^\circ$
$=\cos(48^\circ+12^\circ)\cos(48^\circ-12^\circ)$ $[\cos(\text{A+B})\cos(\text{A}-\text{B})=\cos^2\text{A}-\sin^2\text{B}]$
$=\cos60^\circ\cos36^\circ$
$=\frac{1}{2}\times\Big(\frac{\sqrt{5}+1}{4}\Big)$
$=\frac{\sqrt{5}+1}{8}$
Hence, the correct answer is option A.
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