Question

prove that sin3x.sin³x+cos3x.cos³x=cos³2x
​

Answer 1

Answer:

A simple way to go about such problems is to substitute a value of x as  it is more easier and quicker way to check rather than the conventional method.

But while making such substitutions you have to be careful such that all the values of x is defined, that is x and 3x and 2x in these cases.

I am substituting x=0

So that LHS= 0+0= 0

RHS=0

LHS= RHS

Hence proved.

Answer 2

Given:

$\bold{ \sin 3x \times \sin^3 x+\cos3 x\times cos^3 x=cos^3 2x}$

To find:

proving

Solution:

Formula:

$\bold{\sin 3x= 3 \sin x - 4 \sin^3 x}\\\\\bold{\cos 3x= 4 \cos^3 x- 3 \cos x}$

Equation:

$\sin 3x \times \sin^3 x+\cos3 x\times cos^3 x=cos^3 2x$

solve L.H.S part:

$\Rightarrow \sin 3x \times \sin^3 x+\cos3 x\times cos^3 x\\\\\Rightarrow (3 \sin x - 4 \sin^3 x) \times \sin^3 x+(4 \cos^3 x- 3 \cos x)\times cos^3 x\\\\\Rightarrow (3 \sin^4 x - 4 \sin^6 x) +(4 \cos^6 x- 3 \cos^4 x)\\\\\Rightarrow (3 \sin^4 x- 3 \cos^4 x - 4 \sin^6 x +4 \cos^6 x)\\\\\Rightarrow 3( \sin^4 x- \cos^4 x) - 4( \sin^6 x - \cos^6 x)\\\\\Rightarrow 3( \sin^4 x- \cos^4 x) - 4[( \sin^2 x)^3 - (\cos^2 x)^3]$

$\Rightarrow 3( \sin^4 x- \cos^4 x) - 4[( 1- \sin^2 x \cos^2 x) ( \sin^2 x -\cos^2 x)]\\\\\Rightarrow ( \cos^2 x- \sin^2 x) [3-4-4\sin^2 x \cos^2 x]\\\\\Rightarrow ( \cos2 x) [1-4\frac{1-cos2x}{2}\frac{1+cos2x}{2}]\\\\\Rightarrow ( \cos2 x) [1-1+cos^2 2x]\\\\\Rightarrow ( \cos2 x) (cos^2 2x)\\\\\Rightarrow (cos^3 2x)$

L.H.S = R.H.S