Question

What is `sin^2 3x+cos^2 3x`?

Answer 1

Please see below.

Explanation:

`sin^2A+cos^2A=1` is an identity and is true for all `A`, including `A=3x` and hence

`sin^2 3x+cos^2 3x=1`

However, let us try is using values of `sin3x` and `cos3x`.

But before this as `sin^2x+cos^2x=1`, squaring it

`sin^4x+cos^4x+2sin^2xcos^2x=1` or `sin^4x+cos^4x=1-2sin^2xcos^2x`

and `sin^6x+cos^6x=1-3sin^2xcos^2x(sin^2x+cos^2x)`

= `1-3sin^2xcos^2x` - as `a^3+b^3=(a+b)^3-3ab(a+b)`

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Now coming to proof as

`sin3x=3sinx-4sin^3x` and `cosx=4cos^3x-3sinx`

Therefore `sin^2 3x+cos^2 3x=(3sinx-4sin^3x)^2+(4cos^3x-3cosx)^2`

= `9sin^2x+16sin^6x-24sin^4x+16cos^6x+9cos^2x-24cos^4x`

= `9(sin^2x+cos^2x)+16(sin^6x+cos^6x)-24(sin^4x+cos^4x)`

= `9xx1+16(1-3sin^2xcos^2x)-24(1-2sin^2xcos^2x)`

= `9+16-48sin^2xcos^2x-24+48sin^2xcos^2x`

= `1`

Answer 2

Start from trig identity:
`sin^2 x + cos^2 x = 1`
In this identity, x is a variable, so we can substitute x by another variable X = 3x. Therefore:
`sin^2 X + cos^2 X = sin^2 (3x) + cos^2 (3x) = 1`