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 ...

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 ...

$\begingroup$ $\cos^2(x) = \cos(x)\times \cos(x)$ 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. $\endgroup$ –

2021年12月6日 · $\int \frac {1}{\cos^2 x}\,dx=\int \sec^2 x=\tan x +c$ based directly on the list of immediate integrals. The other day a student asked me if we can evaluate the integral using a method like integration by substitution or integration by parts. The only 'solution' I found uses the differentiation of quotient working backwards. I.e.

2020年4月26日 · One way is to use the complex definitions of sine and cosine. $$\sin\theta=\frac{e^{i\theta}-e^{-i\theta}}{2i} \\\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2 ...

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If the former $(\cos (\arccos x))^2 = x^2$ of course and $\cos(\cos(\arccos a)) = \cos a$ of course. Like you, I far prefer it to mean $ \cos(\cos (a))$ and took it to mean that at first reading. But clearly this text meant it to be $(\cos (a))^2$. $\endgroup$

2014年10月6日 · How do you prove the following: Pythagorean trigonometric identity. For all $\theta\in[0,2\pi]$ it holds that $$ \sin^2\theta+\cos^2\theta=1.$$ I'm curious to know of the different ways of provin...

2018年6月22日 · I broke up the non homogeneous part $\cos^2(x)$ into $\frac12 -\frac12[\cos(2x)]$ and I set up my trial solution as $$ \begin{split} Y_p &= A +B\sin(2x)+C\cos(2x)\\ Y'_p &= 2B\cos(2x)-2C\sin(2x)\\ Y''_p &= -4B\sin(2x)-4C\cos(2x) \end{split} $$ and after substituting these values in the ODE I don't get anywhere.

2015年4月21日 · $\begingroup$ It may simplify things to recognize that $$\cos^2 x = \frac{1}{2} + \frac{1}{2}\cos(2x)$$ is a Fourier series representation of $\cos^2 x$. $\endgroup$ – user169852 Commented Apr 21, 2015 at 2:59