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Question

Please do not just tell me the answer, please provide helpful hints and hide the answers

Using Complex exponential definitions of sine and cosine, prove $\cos\theta=\cos^2 \theta-\sin^2\theta$

All that I know is the trig identity:

$\cos2\theta=1-2\sin^2\theta$ or $\cos2\theta=2\cos^2\theta-1$

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    $\begingroup$ Try De Moivre's formula. $\endgroup$ – Henricus V. Feb 3 '17 at 2:14
  • $\begingroup$ The body of the question doesn't match the title. $\endgroup$ – dxiv Feb 3 '17 at 2:16
  • $\begingroup$ Those trig identities mentioned at the end are false. The identities you're likely thinking of are $\cos(2\theta)=1-2\sin^2(\theta)=2\cos^2(\theta)-1$. $\endgroup$ – Dave Feb 3 '17 at 2:17
  • $\begingroup$ @Dave typo fixed $\endgroup$ – John Rawls Feb 3 '17 at 2:19
  • $\begingroup$ Well, use exponential notation. $e^{ti}=\cos t +i\sin t $. Try figuring that means $e^{2ti} =e^{ti+ti}= (e^{ti})^2$ implies. That should do it. (Remember $i^2=-1$ $\endgroup$ – fleablood Feb 3 '17 at 2:22
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Hint:

$\cos 2\theta + i \sin 2\theta = e^{2i\theta} =\left(e^{i\theta}\right)^2=(\cos \theta+ i\sin \theta)^2$

Then expand the right side and compare real parts

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$\begin{array}{lcl}\cos(2\theta)+i\sin(2\theta) & = & e^{2i\theta} \\ & = & (e^{i \theta})^2 \\ & = & (\cos\theta+i\sin\theta)^2 \\ & = & (\cos\theta)^2+2i\cos θ\sin θ+i^2(\sin θ)^2 \\ & = & (\cos θ)^2+2i\cos θ\sin θ−(\sin θ)^2\end{array}$

Equating real and imaginary parts gives us:

$\begin{array}{lcl}\cos(2θ) = (\cosθ)^2−(\sinθ)^2 \\ \sin(2θ) = 2\cosθ\sinθ \end{array}$

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  • $\begingroup$ cos2θ+isin2θ= e^2iθ $\endgroup$ – Takahiro Waki Jul 29 '18 at 11:38

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