What is ${sin}^{2} 3 x + {cos}^{2} 3 x$ $sin^2 3x+cos^2 3x$ ?

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What is ${sin}^{2} 3 x + {cos}^{2} 3 x$ $sin^2 3x+cos^2 3x$ ?

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Answer 1 · 2017-04-11T13:10:15

Please see below.

Explanation:

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

${sin}^{2} 3 x + {cos}^{2} 3 x = 1$ $sin^2 3x+cos^2 3x=1$

However, let us try is using values of $sin 3 x$ $sin3x$ and $cos 3 x$ $cos3x$ .

But before this as ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$ , squaring it

${sin}^{4} x + {cos}^{4} x + 2 {sin}^{2} x {cos}^{2} x = 1$ $sin^4x+cos^4x+2sin^2xcos^2x=1$ or ${sin}^{4} x + {cos}^{4} x = 1 - 2 {sin}^{2} x {cos}^{2} x$ $sin^4x+cos^4x=1-2sin^2xcos^2x$

and ${sin}^{6} x + {cos}^{6} x = 1 - 3 {sin}^{2} x {cos}^{2} x ({sin}^{2} x + {cos}^{2} x)$ $sin^6x+cos^6x=1-3sin^2xcos^2x(sin^2x+cos^2x)$

= $1 - 3 {sin}^{2} x {cos}^{2} x$ $1-3sin^2xcos^2x$ - as $a^{3} + b^{3} = {(a + b)}^{3} - 3 a b (a + b)$ $a^3+b^3=(a+b)^3-3ab(a+b)$

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Now coming to proof as

$sin 3 x = 3 sin x - 4 {sin}^{3} x$ $sin3x=3sinx-4sin^3x$ and $cos x = 4 {cos}^{3} x - 3 sin x$ $cosx=4cos^3x-3sinx$

Therefore ${sin}^{2} 3 x + {cos}^{2} 3 x = {(3 sin x - 4 {sin}^{3} x)}^{2} + {(4 {cos}^{3} x - 3 cos x)}^{2}$ $sin^2 3x+cos^2 3x=(3sinx-4sin^3x)^2+(4cos^3x-3cosx)^2$

= $9 {sin}^{2} x + 16 {sin}^{6} x - 24 {sin}^{4} x + 16 {cos}^{6} x + 9 {cos}^{2} x - 24 {cos}^{4} x$ $9sin^2x+16sin^6x-24sin^4x+16cos^6x+9cos^2x-24cos^4x$

= $9 ({sin}^{2} x + {cos}^{2} x) + 16 ({sin}^{6} x + {cos}^{6} x) - 24 ({sin}^{4} x + {cos}^{4} x)$ $9(sin^2x+cos^2x)+16(sin^6x+cos^6x)-24(sin^4x+cos^4x)$

= $9 \times 1 + 16 (1 - 3 {sin}^{2} x {cos}^{2} x) - 24 (1 - 2 {sin}^{2} x {cos}^{2} x)$ $9xx1+16(1-3sin^2xcos^2x)-24(1-2sin^2xcos^2x)$

= $9 + 16 - 48 {sin}^{2} x {cos}^{2} x - 24 + 48 {sin}^{2} x {cos}^{2} x$ $9+16-48sin^2xcos^2x-24+48sin^2xcos^2x$

= $1$ $1$

Answer 2 · 2017-04-11T13:27:29

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

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