Very important in integrating things like $\int \cos^{2}(\theta) d\theta$ but it is hard for me to remember them. So how do you deduce this type of formulae? If I can remember right, there was some $e^{\theta i}=\cos(\theta)+i \sin(\theta)$ trick where you took $e^{2 i \theta}$ and $e^{-2 i \theta}$. While I am drafting, I want your ways to remember/deduce things (hopefully fast).

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  1. About TPV's suggestion, how do you attack it geometrically??

$\cos^{2}(x) - \sin^{2}(x)=\cos(2x)$

plus $2\sin^{2}(x)$, then


and now solve homogenous equations such that LHS=A and RHS=B, where $A\in\mathbb{R}$ and $B\in\mathbb{R}$. What can we deduce from their solutions?


In my experience, almost all trigonometric identities can be obtained by knowing a few values of $\sin x$ and $\cos x$, that $\sin x$ is odd and $\cos x$ is even, and the addition formulas: \begin{align*} \sin(\alpha+\beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta,\\ \cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta. \end{align*} For example, to obtain the classic $\sin^2x + \cos^2x = 1$, simply set $\beta=-\alpha$ in the formula for the cosine, and use the facts that $\cos(0)=1$ and that $\sin(-a)=-\sin(a)$ for all $a$.

For the one you have, we use the formula for the cosine with $\alpha=\beta=\theta$ to get $$\cos(2\theta) = \cos^2\theta - \sin^2\theta.$$ Then using $\sin^2 \theta+\cos^2\theta=1$, so $-\sin^2\theta = \cos^2\theta-1$ gives $$\cos(2\theta) = \cos^2\theta +\cos^2\theta - 1 = 2\cos^2\theta - 1.$$

If you know the basic values (at $\theta=0$, $\frac{\pi}{6}$, $\frac{\pi}{4}$, $\frac{\pi}{3}$, $\frac{\pi}{2}$, $\pi$, $\frac{3\pi}{2}$), parity, and the addition formulas, you can get almost any of the formulas.

  • $\begingroup$ I like this way because it feels more general, considering parity. But how do you remember/deduce the addition formulae? What are they? $\endgroup$ – hhh Feb 1 '11 at 14:27
  • $\begingroup$ You can sum them up in in a matrix, that is ctually rather easy to remember [1]. [1] mathworld.wolfram.com/TrigonometricAdditionFormulas.html $\endgroup$ – hhh Feb 1 '11 at 14:29
  • 1
    $\begingroup$ @hhh: The addition formulas are the ones I listed above: the formulas for $\sin(\alpha+\beta)$ and for $\cos(\alpha+\beta)$ in terms of the sines and cosines of $\alpha$ and $\beta$. As to how I remember them, I memorized them. Some things you need to memorize, I just like memorizing as little as possible; my experience is that these two almost always suffice, instead of trying to remember all sorts of other formulas. But the matrix version is certainly a good way of remembering those two. $\endgroup$ – Arturo Magidin Feb 1 '11 at 14:29
  • $\begingroup$ Arturo Magidin: I acccept your question because it lead me to the right solution about addition formulae in matrix notation, I think there must be some name for this type of structures (not just matrices but something more general). $\endgroup$ – hhh Feb 1 '11 at 14:38
  • $\begingroup$ @hhh: What you are seeing with the matrices is simply composition of linear transformations. Don't know what kind of "structure" you are refering to, though, when you talk about "something more general" or "this type of structures". $\endgroup$ – Arturo Magidin Feb 1 '11 at 14:41

By far, the best approach to trigonometric formulae is to use Euler's identity and its corollary, de Moivre's formula, recalling that $\sin^2 + \cos^2 = 1$.


It might be useful to remember that $\cos^2 x$ oscillates twice as fast as $\cos x$. This is something that people who work with alternating current know very well; the effect (which is proportional to the square of the current) has twice the frequency. For example, a light bulb flickers at 100 Hz if the AC frequency is 50 Hz. This means that $\cos^2 x$ should be "something with $\cos 2x$".

Next, since $\cos x$ oscillates between -1 and 1, $\cos^2 x$ will oscillate between 0 and 1. The average around which the curve oscillates will be 1/2, and the amplitude will also be 1/2 (so that you reach down to 0 and up to 1 from the central level 1/2). Thus $\cos^2 x$ should be "1/2 + (1/2)*oscillating term".

Combining these two facts, it's not too hard to remember that $$ \cos^2 x = \frac12 + \frac12 \cos 2x.$$ One has to be a little bit careful with the sign before the second term, but it must be plus if the formula is to hold when $x=0$.

If you choose the minus sign, you get the related formula $$ \sin^2 x = \frac12 - \frac12 \cos 2x.$$ (So $\sin^2 x$ also oscillates around 1/2 with amplitude 1/2 and twice the frequency. Note that when you add the two formulas up, the oscillations cancel, and you get $\cos^2 x + \sin^2 x = 1/2+1/2 = 1$.)

  • $\begingroup$ +1: Great explanation. I love explanations relate to the intuition. $\endgroup$ – Tpofofn Feb 2 '11 at 12:41

For Proving $\sin(\alpha+\beta)=\sin\alpha\cdot \cos\beta + \cos\alpha \cdot \sin\beta$ you can see this link:

By the above by substituting $\beta=\alpha$ you have

  • $\sin{2\alpha} = \sin\alpha \cdot \cos\alpha + \cos\alpha \cdot \sin\alpha =2 \sin\alpha \cos\alpha$

  • $\sin{2\alpha} = \frac{2 \tan\alpha}{1+\tan^{2}\alpha} = \displaystyle 2\tan{\alpha}\cdot\cos^{2}\alpha=2\sin\alpha \cdot \cos\alpha$

We have $\cos(\alpha + \beta) = \cos\alpha\cdot \cos\beta - \sin\alpha \cdot \sin\beta$. Using this you can obtain the following:

  • $\cos{2 \alpha} = \cos\alpha\cdot \cos\alpha - \sin\alpha\cdot \sin \alpha =\cos^{2} \alpha - \sin^{2} \alpha$

  • $\cos{2 \alpha}= 1 - 2 \sin^{2}\alpha \qquad \Bigl[ \because \ \sin^{2}\theta = 1 - \cos^{2}\theta \Bigr]$

  • $ \cos{2 \alpha}=2 \cos^{2}\alpha - 1$

  • $\cos{2 \alpha}=\displaystyle \cos^{2}\alpha \cdot \Bigl[1-\frac{\sin^{2}\alpha}{\cos^{2}\alpha}\Bigr]=\displaystyle \frac{1-\tan^{2}\alpha}{1+\tan^{2}\alpha} \ \Bigl[\because \ \cos^{2}\alpha = \frac{1}{\sec^{2}\alpha}\Bigr]$.

Using the above formulas for $\sin(\alpha + \beta)$ and $\cos(\alpha + \beta)$ we have $$\tan(\alpha+\beta) = \frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)} = \frac{\sin\alpha\cdot \cos\beta + \cos\alpha \cdot \sin\beta}{\cos\alpha\cdot \cos\beta - \sin\alpha\cdot \sin\beta}$$

Dividing the last term by $\cos\alpha \cdot \cos\beta$ gives, $$\tan(\alpha + \beta) = \frac{\tan\alpha + \tan \beta}{1 - \tan\alpha \cdot \tan\beta}$$

  • From this if you put $\beta = \alpha$ you get $$\tan{2\alpha} = \frac{\tan\alpha + \tan\alpha}{1 - \tan\alpha \cdot \tan\alpha} = \frac{2 \tan \alpha}{1 - \tan^{2}\alpha}$$

Now moving on to formulas for $3\alpha$.

\begin{align*} \sin{3\alpha} &= \sin(2\alpha + \alpha) = \sin(2\alpha)\cdot\cos{\alpha} + \cos(2\alpha)\cdot\sin\alpha \\ &= 2\sin\alpha\cdot \cos^{2}\alpha + (\cos^{2}\alpha - \sin^{2}\alpha) \cdot \sin\alpha \\ &= 2\sin\alpha \cdot (1-\sin^{2}\alpha) + (1-\sin^{2}\alpha)\sin\alpha - \sin^{3}\alpha \\ &= 3\sin{\alpha} - 4 \sin^{3}\alpha \end{align*}

Again proceed similarly for $\cos{3\alpha}$ and convert all the $\sin^{2}\alpha$'s which appear in the expression to $(1-\sin^{2}\alpha)$. Then you get the value as:

  • $\cos(3\alpha) = 4\cos^{3}\alpha - 3 \cos\alpha$.

  • $\displaystyle \tan{3\alpha} = \frac{3\tan\alpha - \tan^{2}\alpha}{1-3\tan^{2}\alpha}$. Look here

  • $\begingroup$ @hhh: Hope this helps. $\endgroup$ – user9413 May 18 '11 at 19:18

For this you need only two very basic trigonometric identities: $$ \cos^2(x) + \sin^2(x) = 1 $$ and $$ \cos^2(x) - \sin^2(x) = \cos(2x) $$ Add or subtract them, and you have the identities for $\cos^2(x)$ and $\sin^2(x)$.

  • $\begingroup$ Problem with the last is how you remember it or deduce it fast. I cannot see it directly. $\endgroup$ – hhh Feb 1 '11 at 12:52
  • $\begingroup$ I think in this case (2x) it's easier just to know it by heart than to use the Moivre's formula, but it's your call :). I had to learn some basic trigonometric in high school like $\sin(A+B)$ and $\cos(A+B)$, so that's what naturally comes to my mind. About the transformation: idk if that helps or not, for me a $-2\sin^2(x)$ isn't much of a help or starting point. $\endgroup$ – tpv Feb 1 '11 at 13:01

All of trig identities can be derived from $\exp{(\hat{i}\theta)}=\cos(\theta)+\hat{i}\sin\theta$. How?


$$\cos(2\,x) = \mathrm{Re}\left(\exp{(2\hat{i}\,x)}\right)$$ $$ = \mathrm{Re}\left(\exp{(\hat{i}x)}\,\exp{(\hat{i}x)}\right) $$ $$ = \mathrm{Re}\left((\cos x+\hat{i}\sin x)\,(\cos x+\hat{i}\sin x)\right) $$ $$ = \mathrm{Re}\left( \cos^2 x - \sin^2 x +2\hat{i}\cos x \sin x \right) $$ $$ = \cos^2 x - \sin^2 x $$


Here's yet another method for your amusement.

Let $y = \cos^2(t)$. Then note that $y'(t) = 2\cos(t)\sin(t)$ and therefore $$ y''(t) = 2\cos^2(t) - 2\sin^2(t) = 2\cos^2(t) - 2(1 - \cos^2(t)) = 4 y - 2. $$ The general solution to this nonhomogeneous ode is $$ y = C_1 \sin(2t) + C_2 \cos(2t) + 1/2 $$ and since we know that $y$ satisfies the initial conditions $y(0) = 1$ and $y'(0) = 0$, we can solve for the coefficients and get $C_1 = 0$ and $C_2 = 1/2$.


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