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

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 ...

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.

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

2017年7月30日 · I'll need to memorize $\cos2x = \cos^2x - \sin^2x$ as I'll use it in derivatives. Only, there are other forms for this identity, I can't see how I can get to the others from this one above. The other forms are. $$2\cos^2x - 1\\ 1 - 2\sin^2x$$

2016年10月7日 · The question is what is the minimum value of $$2^{\sin^2x}+2^{\cos^2x}$$ I think if I put $x=\frac\pi4$ then I get a minimum of $2\sqrt2$.

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We know the Maclaurin series for #cos(x)#: #sum_(n=0)^oox^(2n)/((2n)!)*(-1)^n=1-x^2/(2!)+x^4/(4!)-x^6/(6!)...# ...

2015年8月25日 · Solve f(x) = cos^2 x + 2cos 2x + 1 = 0 Ans: +- 63.43 and +- 116.57 deg Replace in the equation cos 2x by (2cos^2 x - 1) f(x) = cos^2 x + 4cos^2 x - 2 + 1 = 0 5cos^2 x - 1 = cos^2 x = 1/5 --> cos x = +- 1/sqrt5 = +- 0.48 a. cos x = 0.48 --> x = +- 63.43 deg b. x = - 0.48 --> x = +- 116.57 Check by calculator: x = 63.43 deg --> cos^2 x = 0.20 ; 2cos 2x = - 1.20. f(x) = 0.20 - 1.20 + 1 = 0. OK x ...