Truthfully, the notation $\cos^2(x)$ should actually mean $\cos(\cos(x)) = (\cos \circ \cos)(x)$, that is, the 2nd iteration or compositional power of $\cos$ with itself, because on an arbitrary space of self-functions on a given set, the natural "multiplication" operation is composition of those functions, and the power is applied to the ...

$\begingroup$ $\cos^2(x) = \cos(x)\times \cos(x)$ and $\cos(x^2) = \cos(x \times x)$ So no. But beware, the notation $\cos^{-1}(x)$ is ambiguous. It can denote the inverse cosine function or the reciprocal of the cosine function. $\endgroup$ –

the same diagram also gives an easy demonstration of the fact that $$ \sin 2x = 2 \sin x \cos x $$ as @Sawarnak hinted, with the help of this result, you may apply your original idea to use calculus for an easy derivation, since differentiation gives $$ 2 \cos 2x = 2(\cos^2 x - \sin^2 x) $$ it is not a bad idea to familiarize yourself with several different 'proofs' of such fundamental ...

2020年7月29日 · $$\\cos^2(\\theta) + \\sin^2(\\theta) = 1$$ I solved this by using right triangle, $$\\sin(\\theta) = \\frac{a}{c}, \\quad \\cos(\\theta) = \\frac{b}{c}$$ $$\\cos^2 ...

2014年10月6日 · Therefore, $\rm OM=\sqrt{\overline{OC}^2+\overline{OS}^2}=\sqrt{\cos^2\theta+\sin^2\theta}$. Since $\rm M$ lies in the unit circle, $\rm OM$ is the radius of that circle, and by definition, this radius is equal to $1$. It immediately follows that: $$\color{grey}{\boxed{\,\displaystyle\color{black}{\cos^2\theta+\sin^2\theta=1}}}$$ $\phantom{X}$

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$$ \cos^2 i = {(b/a)^2 − (b/a)^2_{eos} \over 1 − (b/a)^2_{eos}} $$ All I need however is to determine ...

$\cdots \sin(-\pi), \sin(0), \sin(\pi), \sin(2\pi), \sin(3\pi),\cdots$ Which is exactly where the sine function has its roots, so it is always equal to $0$. For the cosine case, use the identity $\cos(x) = \cos(x + 2\pi) $ (period of the cosine function is $2\pi$) and plug $\cos(0)$ and $\cos(\pi)$ to …

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I ended up realizing that the easiest way to go about it would be to convert the $ cos^2(2\pi t) $ into a term without a square via the double angle formulas, so that it would turn into $ 1 \over 2 $ * $ cos(4\pi t) $ and some constant.