Why is it that:
$$\sin^2(-x) = \sin^2(x) $$
And:
$$\sec^2(-x) = \sec^2(x)$$
Would the same hold true if the exponent was a $3$ or a $5$? How come the same isn't applicable to sin or the other trigonometric functions?
$\endgroup$ 22 Answers
$\begingroup$Because Sine is an odd function, we have that $\sin(-x)^2 = (-\sin x)^2 = (-1)^2 \sin(x)^2 = \sin(x)^2$
Similar logic yields your second identity
$\endgroup$ $\begingroup$Observe that if suppose $f(x)$ is an even function then $f(x) = f(-x)$,
so $f^{n}(x) = f^{n}(-x)$
and now suppose that $g(x)$ is an odd function
so $g(-x) = -g(x)$
Hence when we take the power (composition) that is $g^{n}(-x) = (-1)^{n}g^{n}(x)$ and thus now the equality between the two is obtained depending upon whether $n$ is odd or even.
If $n$ is even then $g^{n}(-x) = (-1)^{n}g^{n}(x) = g^{n}(x)$ and
for the case of odd $n$,we get $g^{n}(-x) = (-1)^{n}g^{n}(x) = - g^{n}(x)$.
Coming to Trigonometric functions we know $\sin(x)$ is an odd function,and $\cos(x)$ is an even functions and you can apply the above theory now and check for any power now!.
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