Question

If $sin 2 \beta $ is the geometric mean between $sin \alpha $ and $cos \alpha ,$ then $cos 4 \beta $ is equal to

• A. $2\left(sin\right)^{2} \left(\frac{\pi }{4} - \alpha \right)$
• B. $2\left(cos\right)^{2} \left(\frac{\pi }{4} - \alpha \right)$
• C. $2\left(cos\right)^{2} \left(\frac{\pi }{2} + \alpha \right)$
• D. $2\left(sin\right)^{2} \left(\frac{\pi }{4} + \alpha \right)$

Answer 1

Answer: (A)

$sin^{2} 2 \beta =sin ⁡ \alpha cos ⁡ \alpha $
$cos 4 \beta =1-2sin^{2} ⁡ 2 \beta $
$=sin^{2} \alpha +cos^{2} ⁡ \alpha -2sin ⁡ \alpha cos ⁡ \alpha $
$=\left(sin \alpha - cos ⁡ \alpha \right)^{2}$
$=\left(\sqrt{2} \left(sin \alpha \cdot \frac{1}{\sqrt{2}} - cos \alpha \frac{1}{\sqrt{2}}\right)\right)^{2}$
$=2\left(sin \alpha cos ⁡ \frac{\pi }{4} - cos ⁡ \alpha sin ⁡ \frac{\pi }{4}\right)^{2}$
$=2\left(sin \left(\alpha - \frac{\pi }{4}\right)\right)^{2}=2\left(sin\right)^{2} ⁡ \left(\frac{\pi }{4} - \alpha \right)$