Question

2. Express the product of sines and cosines


sin 2 theta + sin 4 theta ​

Answer 1

Answer:

We have,

sin12θ+sin4θ

⇒sin2(6θ)+sin2(2θ)

⇒2sin6θcos6θ+2sin2θcos2θ

⇒2sin2(3θ)cos6θ+4sinθcosθcos2θ

⇒4sin3θcos3θcos6θ+4sinθcosθcos2θ

Step-by-step explanation:

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Answer 2

Answer:

sin 2$\theta$ + sin 4$\theta$ = 6 sin $\theta$ cos³$\theta$ - 2 sin³$\theta$ cos $\theta$

Step-by-step explanation:

Given sin 2$\theta$ + sin 4$\theta$

As sin 2$\theta$ = 2 sin $\theta$ cos $\theta$

=> 2 sin $\theta$ cos $\theta$ + 2 sin 2$\theta$ cos 2$\theta$

= 2 sin $\theta$ cos $\theta$ + 2 * 2 sin $\theta$ cos $\theta$ cos 2$\theta$

Taking common

=> 2 sin $\theta$ cos $\theta$ [1 + 2 cos 2$\theta$]

As cos 2$\theta$ = cos² $\theta$ - sin² $\theta$

=> 2 sin $\theta$ cos $\theta$ [ 1 + 2(cos² $\theta$ - sin² $\theta$) ]

As cos² $\theta$ + sin² $\theta$ = 1

=> 2 sin $\theta$ cos $\theta$ [ cos² $\theta$ + sin² $\theta$  + 2cos² $\theta$ - 2sin² $\theta$ ]

= 2 sin $\theta$ cos $\theta$ [ 3cos² $\theta$ - sin² $\theta$ ]

= 6 sin $\theta$ cos³$\theta$ - 2 sin³$\theta$ cos $\theta$

=> sin 2$\theta$ + sin 4$\theta$ = 6 sin $\theta$ cos³$\theta$ - 2 sin³$\theta$ cos $\theta$