Question

solve this no spam question no 3​

Answer 1

Question :

Express the following as the product of sines and cosines

(3) sin 2θ + cos 4θ

Answer :

Given :

sin 2θ + cos 4θ

Using trigonometric ratios of complementary angles, cos Φ = sin ( 90 - Φ ) we get

⇒ sin 2θ + sin ( 90 - 4θ )

Applying Sum - to - Product identity we get,

$\boxed{ \rm sin \ C + sin \ D = 2sin\dfrac{C + D}{2}.cos\dfrac{C - D}{2}}$

$\Rightarrow \sf 2sin\dfrac{2 \theta + (90 - 4 \theta)}{2}.cos\dfrac{2 \theta - (90 - 4 \theta)}{2}$

$\Rightarrow \sf 2sin\dfrac{90 - 2\theta}{2}.cos\dfrac{2 \theta - 90 + 4 \theta}{2}$

$\Rightarrow \sf 2sin\dfrac{90 - 2\theta}{2}.cos\dfrac{6 \theta - 90 }{2}$

$\Rightarrow \sf 2sin\dfrac{2(45 - \theta)}{2}.cos\dfrac{2(3 \theta - 45) }{2}$

$\Rightarrow \sf 2sin \ (45 - \theta).cos \ (3 \theta - 45)$

It can be written as

$\Rightarrow \sf 2sin \ (45 - \theta).cos \ \{ - (45 - 3 \theta ) \}$

Since cos ( - Φ ) = cos Φ

$\Rightarrow \sf 2sin \ (45 - \theta).cos \ (45 - 3 \theta)$

Hence expressed !

Answer 2

↑↑you answer refer to the attachment↑↑