Question

Prove the Identity.

sinx[sinx+sin(5x)]=cos(2x)[cos(2x)-cos(4x)]

Answer 1

The Identities I am using are:

\[\large sinA + sinB = 2 \cdot \sin(\frac{A + B}{2})\cos(\frac{A - B}{2})\]

\[\large 2sinAsinB = \cos(A-B) - \cos(A+B)\]

Answer 2

So solve the brackets first using first Identity I have given above:

\[\sin(x)(\sin(x) + \sin(5x)) \implies \sin(x)( \color{green}{2 \cdot \sin(3x) \cdot \cos(2x))}\]

\[\implies \color{blue} {(2 \cdot \sin(3x) \cdot \sin(x))} \cdot \cos(2x) \implies \color{cyan}{(\cos(2x) - \cos(4x))} \cdot \cos(2x)\]

Rearrange them to get equal to RHS:

\[\implies \color{red}{\cos(2x) } \color {darkgreen} {(\cos(2x) - \cos(4x))}\]

Answer 3

@rosiepose99 go through the steps and tell me do you have any problem in any step??

Answer 4

ok can you explain how you did the final step where you had (2*sin(3x)*sin(x))*cos (2x)

Answer 5

why did you put the extra sin(x) before the cos(2x)

Answer 6

It is not the extra sin(x) see you have sin(x) in outside too..

So I have just taken that backwards..

Answer 7

in mutiplication you can change the preference order:

see:

\[\large 5 \times 4 \times 2 \implies 4 \times 5 \times 2 \implies 2 \times 4 \times 5\]

They all are one and same thing..

Answer 8

So:

\[\sin(x) \times 2 \times \sin(3x) \times \cos(2x)\]

I have written this as:

\[2 \times \sin(3x) \times \sin(x) \times \cos(2x)\]

Getting??

Answer 9

I am expecting a reply from you @rosiepose99

Answer 10

sorry and thank you, I understand it.