# Could someone please explain double-angle identities?

I don't understand how to do maths, mostly because I don't understand why formulae work they way they do, or the reasoning behind equations, etc.

I tried to explain the $\sin(2\theta)$ double-angle identity to myself but failed:

Hypothetically if:

$$\text{opp} = 1 \qquad \text{adj} = 2 \qquad \text{hyp} = 3$$

then

\begin{align*} \sin(2\theta) &= 2\sin(\theta)\cos(\theta)\\\\ \left(\frac{\text{opp}}{\text{hyp}}\right)\cdot 2 &= 2\cdot\left(\frac{\text{opp}}{\text{hyp}}\right)\left(\frac{\text{adj}}{\text{hyp}}\right)\\\\ \frac{1}{3}\cdot 2 & = 2\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)\\\\ \frac{2}{3} &\neq \frac{4}{9} \end{align*}

Where did I go wrong? How do the double-angle identities work?

• $\sin(2\theta)$ does not equal $\sin(\theta)\times2$! – Nick D. May 28 '15 at 4:33
• Alright, I'll keep this in mind. Thank you for helping! – Vesta May 28 '15 at 5:51

The best way to see how the identities work is to see WHY they work. I find that formulas are much more illuminating when one sees a proof. This will, in trigonometry, usually appeal to some geometric intuition while giving you a general formula. So, we can try to prove an identity such as $$\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a).$$ Then, using this general formula, what would we know about $\sin(2a)=\sin(a+a)$?

For a proof of the general angle sum formula, here is a fairly nice geometric approach which you may find illuminating.

• Yes! This was exactly what I needed and I will definitely read this! Thank you! – Vesta May 28 '15 at 5:47

Also, a right triangle with those sides does not exist due to Pythagorean Theorem.

If you need to check the formula then take for instance $\theta = \pi /2$. Then $\sin(2\theta) = \sin(\pi) = 0$. On the other hand, $2\sin(\theta)\cos(\theta)= 2 \cdot 1 \cdot 0=0$.

• Alright, but could you explain how the identities work, preferably using mathematical examples? Thanks! – Vesta May 28 '15 at 4:48
• O' my brain finally caught on with what you said facepalms.. Thanks for your help! – Vesta May 28 '15 at 5:45

There are two problems with your calculation:

1. The trio of opposite, adjacent, and hypotenuse refer to a right-angled triangle. A triangle is right-angled if and only if it satisfies $a^2 + b^2 = c^2$ (or, if you like, $opp^2 + adj^2 = hyp^2$), which yours does not. In other words, you took as an example a lop-sided triangle, and it isn't really meaningful to talk about the "hypotenuse" anymore.

2. In your calculation, you wrote that $$\sin(2\theta) = 2\big(\frac{opp}{hyp}\big).$$ But you can't pull out the $2$: it's not true that $\sin(2\theta) = 2\sin\theta$. It would be more appropriate to write $$\sin(2\theta) = \frac{opp_2}{hyp_2}$$ where $opp_2$ and $hyp_2$ are the opposite side and hypotenuse in a different triangle.

As an example, let's take $\theta = 30^\circ$, so we have a $30^\circ-60^\circ-90^\circ$ triangle. The side lengths are: $$adj_1 = \frac{\sqrt3}{2} \quad opp_1 = \frac12 \quad hyp_1 = 1.$$ We have $\sin\theta = \frac{opp_1}{hyp_1} = \frac12$, and $\cos\theta = \frac{adj_1}{hyp_1} = \frac{\sqrt3}{2}$.

Now, $2\theta = 60^\circ$, in which case the side lengths are: $$adj_2 = \frac12 \quad opp_2 = \frac{\sqrt3}{2} \quad hyp_2 = 1.$$ (It's also a $30^\circ-60^\circ-90^\circ$ triangle, but flipped the other way.) For this triangle, where the angle of interest is $2\theta$, we have $\sin(2\theta) = \frac{opp_2}{hyp_2} = \frac{\sqrt3}{2}$.

You can see that indeed $\sin(2\theta) = 2\sin\theta\cos\theta$, because $$\frac{\sqrt3}{2} = 2 \big(\frac12\big)\big(\frac{\sqrt3}{2}\big).$$

• Alright, it makes a lot more sense now! I tried it again using proper right triangle measurements and it checked! I can't believe I didn't catch on with that sooner facepalms. Thanks you! – Vesta May 28 '15 at 5:44