I was studying math.. and I just realized that I only just memorized these trigonometric equations, but I don't really know the reason behind them. So um...

Why is $\cos(2x)=\cos^2(x)-\sin^2(x)$ and $\sin(2x)=2\sin(x)\cos(x)$?

What happens if $x=3\theta$? Would the equations change to something like

$\cos(6\theta)=\cos^2(3\theta)-\sin^2(3\theta)$ and


  • $\begingroup$ These are both immediate from the addition formulas for sine and cosine, which you can see proved here. In response to your second question, yes, both formulas are correct. $\endgroup$ – user61527 Jul 5 '14 at 4:15
  • $\begingroup$ Yes, precisely. You may further reduce $\cos(3\theta)$ and $\sin(3 \theta)$ using other identities until you reach $\cos (\theta)$ and $\sin (\theta)$ or whatever else you may need. $\endgroup$ – Mark Fantini Jul 5 '14 at 4:16
  • $\begingroup$ With a little effort you can find, and test (maybe by induction) , a formula for $cos(n \theta)$ and for $sin(n\theta)$ using, e.g., Bonger's advice. $\endgroup$ – user99680 Jul 5 '14 at 4:21
  • $\begingroup$ Thank you everyone!!@bongers, i will study that now! $\endgroup$ – user125342 Jul 5 '14 at 4:29
  • $\begingroup$ Did you really mean to ask about the case $x = 3\theta$? That doesn't offer much insight, as you can see. I'm thinking you wanted to ask about $\cos 3x$ and $\sin 3x$. $\endgroup$ – M. Vinay Jul 5 '14 at 6:21

One way to see and also remember easier the always evolving trigonometric identities is using (a representation of) complex numbers and DeMoivre's theorem (or more correctly Euler's identity) like this:

$$e^{ix} = \cos(x) + i\sin(x)$$


$$e^{i2x} = e^{ix} \times e^{ix}$$

$$\cos(2x) + i\sin(2x) = (\cos(x) + i\sin(x)) \times (\cos(x) + i\sin(x))$$

by equating real and imaginary parts respectively one gets the identities and so on..

Now this is purely formal as stated, but by actually taking into account the geometrical meaning (or representation) of a complex number (as is assumed in Euler's representation) this is made way more intuitive. For example multiplication of 2 complex numbers (of modulus $1$ i.e $e^{ia}$) has a direct meaning as geometrical rotation (on the unit circle) and the rest follow from that.

Furthermore, this representation of complex numbers of modulus one as representing rotations is further used in groups like $SU(n)$ and $U(1)$ which play a central part in physics and engineering (for example Quantum Mechanics)


$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$ Proof without words:


$\qquad\quad$ Image credit: "Blue's Blog: The Bloog!". See also this Math.SE answer by Blue.

  • $\begingroup$ Nice, based on pythagorean theorem directly (as indeed trigonometry is based on the pythagorean theorem). In my answer the pythagorean theorem is encoded inside the complex DeMoivre representation $\endgroup$ – Nikos M. Jul 5 '14 at 8:14
  • $\begingroup$ @Nikos: Where is Pythagoras here? $\endgroup$ – TonyK Jul 5 '14 at 9:25
  • $\begingroup$ Very nice - an image is worth more than 1000 words!!!! $\endgroup$ – johannesvalks Jul 5 '14 at 9:26
  • $\begingroup$ @TonyK, the pythagorean relation is first of all in the fact that $1 = (\cos \beta)^2 + (\sin \beta)^2$, the other right-angled triangles have same relations. Then by parallels and angles similarity the rest relations follow. Also $\cos \alpha = adjacent side / hypotenuse \implies adjacent side = \cos \alpha \times \cos \beta$ etc... (which is also related to pythagorean theorem) $\endgroup$ – Nikos M. Jul 5 '14 at 9:33
  • $\begingroup$ This should be the best answer... $\endgroup$ – johannesvalks Jul 5 '14 at 9:37

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