Note for trigonometric functions, $\cos^{-1}$ sometimes refers to $\arccos$, and sometimes to $\sec = \frac1{\cos}$, so you should be careful about exponentiating functions. Share Cite

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 ...

Feb 24, 2015 · Because $\cos{2x}$ is already an even function when I extend it to $(-\pi, \pi)$, I felt comfortable just ...

Otherwise, you can use the Pythagorean Identity $\sin^2x + \cos^2x = 1$ to solve for $\cos x$, remembering to take the negative root. Once you find $\cos x$, substitute the values of $\sin x$ and $\cos x$ into the double angle formulas for $\sin(2x)$ and $\cos(2x)$, from which you can determine the values of the other trigonometric functions.

Mar 24, 2012 · The key here is the Pythagorean identity, which states that $\sin^2x+\cos^2x=1$. You may have seen the ...

Dec 20, 2015 · It's a Pythagorean identity and comes from $$\sin^2 x + \cos ^2 x = 1$$ Just subtract $\cos ^2 x$ from both sides and you have your answer.

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Apr 21, 2015 · $\begingroup$ 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

Now, you need to expand these brackets and follow the same procedure to simplify $\cos x \cos x$, $\cos x \cos 7x$, $\cos 3x \cos x$ and $\cos 3x \cos 7x$. Share Cite

For both expressions to be zero we must have $$\sin(2x) = -\sin(2x+y) = \sin(y).$$ For the first expression to equal the last we must have that $$\sin(2x) - \sin(y) = 2\sin({2x - y \over 2}) \cos({2x + y\over 2}) = 0$$ where we use a trig identity for the difference of sines. So either $2x = y+2m\pi $ or $ 2x = -y + (2m+1)\pi $ for some integer ...