Question

Please help with number 7

Answer 1

Answer:

Step-by-step explanation:

cos 2α=1-2 sin² α

$sin \alpha =\sqrt{\frac{1-cos2\alpha }{2} } \\put~ \alpha =\frac{\pi }{8} \\2\alpha =\frac{\pi }{4} \\cos 2\alpha =cos \frac{\pi }{4}=\frac{1}{\sqrt{2} } \\\\sin(\pi/8)=\sqrt{\frac{1-\frac{1}{\sqrt{2} } }{2}} =\sqrt{\frac{\sqrt{2}-1 }{2\sqrt{2} }$

6 sin 3 \theta+6 sin \theta

$6 sin 3 \alpha +6 sin\alpha =6(sin 3\alpha +sin \alpha )\\=6 *2sin \frac{3\alpha+\alpha }{2} cos \frac{3\alpha-\alpha }{2} \\=12 sin 2\alpha cos \alpha$

please change α to \theta