What is \(\sin ^{2} 3 x+\cos ^{2} 3 x ?\)
Answer 1:
Please see below.
Explanation:
\(\sin ^{2} A+\cos ^{2} A=1\) is an identity and is true for all \(A\), including \(A=3 x\) and hence
However, let us try is using values of \(\sin 3 x\) and \(\cos 3 x\).
But before this as \(\sin ^{2} x+\cos ^{2} x=1\), squaring it
\(\sin ^{4} x+\cos ^{4} x+2 \sin ^{2} x \cos ^{2} x=1 \text { or } \sin ^{4} x+\cos ^{4} x=1-2 \sin ^{2} x \cos ^{2} x\)
and \(\sin ^{6} x+\cos ^{6} x=1-3 \sin ^{2} x \cos ^{2} x\left(\sin ^{2} x+\cos ^{2} x\right)\)
\(=1-3 \sin ^{2} x \cos ^{2} x-\text { as } a^{3}+b^{3}=(a+b)^{3}-3 a b(a+b)\)
Now coming to proof as
\(\sin 3 x=3 \sin x-4 \sin ^{3} x\) and \(\cos x=4 \cos ^{3} x-3 \sin x\)
Therefore \(\sin ^{2} 3 x+\cos ^{2} 3 x=\left(3 \sin x-4 \sin ^{3} x\right)^{2}+\left(4 \cos ^{3} x-3 \cos x\right)^{2}\)
\(=9 \sin ^{2} x+16 \sin ^{6} x-24 \sin ^{4} x+16 \cos ^{6} x+9 \cos ^{2} x-24 \cos ^{4} x\)
\(=9\left(\sin ^{2} x+\cos ^{2} x\right)+16\left(\sin ^{6} x+\cos ^{6} x\right)-24\left(\sin ^{4} x+\cos ^{4} x\right)\)
\(=9+16-48 \sin ^{2} x \cos ^{2} x-24+48 \sin ^{2} x \cos ^{2} x\)
\(=1\)
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:
\(\sin ^{2} X+\cos ^{2} X=\sin ^{2}(3 x)+\cos ^{2}(3 x)=1\)