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I was hoping someone could explain how $\cos^2 x-\sin^2 x = \cos 2x$

After using the product rule to differentiate $\sin x \cdot \cos x$ I get the answer $\cos^2 x - \sin ^2 x$ I've come across this problem twice now and each time I've gotten the same wrong so I'm hoping someone can point out what I'm missing.

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    $\begingroup$ Are you sure it's $\cos^2x$ instead of $\cos2x$? $\endgroup$ – Joonas Ilmavirta Aug 24 '14 at 11:12
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    $\begingroup$ Probably a typesetting error, or a parsing error, $\cos^2 x - \sin^2 x = \cos (2x)$. $\endgroup$ – Daniel Fischer Aug 24 '14 at 11:12
  • $\begingroup$ Yes sorry you are right, it should = cos(2x) $\endgroup$ – Paul Aug 24 '14 at 11:15
  • $\begingroup$ Do you know that $2\sin x\cos x=\sin (2x)$? $\endgroup$ – Sawarnik Aug 24 '14 at 11:54
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$$\cos(2x) \equiv \underbrace{\cos(\color{green}{x}+\color{red}x) \equiv \cos(\color{green}x)\cos(\color{red}x)-\sin(\color{green}x)\sin(\color{red}x)}_{\text{addition identity for} \ \ \cos(\color{green}\alpha+\color{red}\beta)} \equiv \cos^2(x)-\sin^2(x).$$

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  • $\begingroup$ So it just needed simplified by a trigonometric identify? It all seems so simple once you know how, thank you for clearing this up $\endgroup$ – Paul Aug 24 '14 at 11:20
  • $\begingroup$ @Paul: Very often, it seems that the hardest part of Calculus is the Pre-Calculus. :) $\endgroup$ – Blue Aug 24 '14 at 11:48
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    $\begingroup$ Why use $\equiv$ instead of $=$ ? $\endgroup$ – Nick Aug 24 '14 at 12:49
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    $\begingroup$ @Nick The $\equiv$ symbol, in this context, means "is equal for all values of $x$" (as opposed to $=$ which is usually true for only some specific values of $x$). e.g. $x(x+1) \equiv x^2+1$ (these two expressions are algebraically equivalent-- they're equal for all $x$), vs. $2x=4$ which is true only for one value of $x$ (namely $x=2$). $\endgroup$ – beep-boop Aug 24 '14 at 13:25
  • $\begingroup$ @alexqwx: So, should I be using $\equiv$ in statements like $$6x\times 4 \equiv 3x \times 8$$ $\endgroup$ – Nick Aug 24 '14 at 13:33
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$\cos^2 (x)-\sin^2(x)=\cos(x)\cos(x)-\sin(x)\sin(x)= \cos(x+x)=\cos(2x)$

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Algebraic proof. $$\cos(2x)=\cos(x+x)=\cos x \cos x - \sin x \sin x = \cos^2 x - \sin^2 x$$

Using de Moivre's formula. $$\cos(2x) + i \sin(2x) = (\cos x + i \sin x)^2 = \\ \cos^2 x + i^2 \sin^2 x + 2i \cos x \sin x = \cos^2 x - \sin^2 x + 2i \cos x \sin x .$$ From here we get $\cos(2x) = \cos^2 x - \sin^2 x.$

You can find more hints at ProofWiki.

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It can be proved without using trigonometric identity $\cos2x=\cos^2x-\sin^2x$. $$\cos^2x-\sin^2x=\frac{1+\cos2x}{2}-\frac{1-\cos2x}{2}=\frac{1+\cos2x-1+\cos2x}{2}=\frac{2\cos2x}{2}=\cos2x$$

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let $OPN$ be an isosceles triangle with $OP=ON=1$ and $O\hat PN = x$

let $H$ be the midpoint of $NP$ and let $M$ be the foot of the perpendicular from $P$ to the continuation of $NO$

since the right-angled triangles $NHO$ and $NMP$ are similar we have: $$ \frac{NP}{NM} = \frac{ON}{HN} $$ now, using simple geometry and elementary trig on right-angled triangles we have $$ \begin{align} HN &= \cos x \\ ON &= 1\\ NP &= 2\cos x\\ NM &=1+\cos 2x \end{align} $$ thus $$ \frac{2\cos x}{1 +\cos 2x} = \frac1{\cos x} $$ or $$ \cos 2x = 2 \cos^2 x - 1 $$ but for all $x$, $$ 1 = \cos^2x + \sin^2 x $$ giving: $$ \cos 2x = 2 \cos^2 x - ( \cos^2x + \sin^2 x ) $$ and the required result immediately follows

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  • $\begingroup$ :Would you please tell me how $MN=1+\cos 2x$ in $$\begin{align} HN &= \cos x \\ ON &= 1\\ NP &= 2\cos x\\ NM &=1+\cos 2x \end{align}$$ ? $\endgroup$ – Khosrotash Dec 24 '17 at 19:06
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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 identities. this helps to clarify the different assumptions required, and leads to insights about the relationship between different mathematical idioms.

for example if $z$ is a complex number we have: $$ \mathfrak{Re}(z^2) = \mathfrak{Re}(z)^2 - \mathfrak{Im}(z)^2 $$ and this leads to a slightly different demonstration resting on the identity: $$ \mathfrak{arg}(z^2) = 2 \mathfrak{arg}(z) $$

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