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

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

$\sin(6\theta)=2\sin(3\theta)\cos(3\theta)$?

• 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.
– user61527
Commented Jul 5, 2014 at 4:15
• 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. Commented Jul 5, 2014 at 4:16
• 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. Commented Jul 5, 2014 at 4:21
• Thank you everyone!!@bongers, i will study that now! Commented Jul 5, 2014 at 4:29
• 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 $\sin 3x$. Commented Jul 5, 2014 at 6:21

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

$$\qquad\quad$$

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

• 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 Commented Jul 5, 2014 at 8:14
• @Nikos: Where is Pythagoras here? Commented Jul 5, 2014 at 9:25
• Very nice - an image is worth more than 1000 words!!!! Commented Jul 5, 2014 at 9:26
• @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) Commented Jul 5, 2014 at 9:33
• This should be the best answer... Commented Jul 5, 2014 at 9:37

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:

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

Then

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

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

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)

• Sorry for the late comment, but is your first statement Euler's identity, and the next statement based on De Moivre's? Commented Jun 18 at 10:19
• I use $e^{ix} = \cos(x) + i\sin(x)$ which is Euler's formula (DeMoivre's formula can be seen as special case of that) Commented Jun 18 at 15:07