# Prove $(1-\cos x)/\sin x = \tan x/2$

Using double angle and compound angles formulae prove,

$$\frac{1-\cos x}{\sin x} = \tan\frac{x}{2}$$

• Write it as $(1 - \cos(2y))/\sin(2y) = \tan y$. Do you see then how to proceed? – Daniel Fischer Jul 10 '13 at 13:39
• I'll try proceeding, thanks for the tip! – Red Queen10101 Jul 10 '13 at 13:42
• Im revising for a test, and im stuck on this question, and I still don't understand many of the hints/solutions people have given me. – Red Queen10101 Jul 10 '13 at 14:28

$$\dfrac{1-\cos x}{\sin x}=\dfrac{1-(1-2\sin^2\frac x2)}{2\sin\frac x 2\cos\frac x2}=\dfrac{\sin\frac x2}{\cos\frac x2}=\tan\frac x2$$

• When a problem is marked "homework" please don't answer the problem completely. – Thomas Andrews Jul 10 '13 at 13:47
• THANKS! makes self teaching way easier – Red Queen10101 Jul 10 '13 at 14:51

Use geometry: $AO= 1$ It is strange that no one has mentioned this drawing yet.

P.S. Note that you can easily extract other trigonometric identites involving $\phi/2, 2\phi$ argument from this picture. For example to get $sin(\phi/2)$ use $ECD$ triangle and Pythagorean theorem to calculate $CD/ED = sin (\phi/2)$

• I think, this is what Red Queen10101 want – ziang chen Jul 10 '13 at 14:24
• we haven't been taught this at school. – Red Queen10101 Jul 10 '13 at 14:26
• School in which pupils did not draw circles and triangles...O tempora, o mores! – igumnov Jul 10 '13 at 14:32

Hint: It is easier to show:

$$\frac{1-\cos 2y}{\sin 2y} = \tan y$$

using the formulas for $\cos 2y$ and $\sin 2y$.

• hi ive used the formulas for cos2y and sin2y. Im still not catching the drift of continuing on. – Red Queen10101 Jul 10 '13 at 13:52
• What did you get? – Thomas Andrews Jul 10 '13 at 14:02
• (1- cos^2 x +sin^2 x)/ (2sinxcosx) – Red Queen10101 Jul 10 '13 at 14:08
• sorry the x represents y – Red Queen10101 Jul 10 '13 at 14:11
• $\cos^2 2x = 1-\sin^2 2x$ – ziang chen Jul 10 '13 at 14:27

$$\cos x = \cos^2\frac{x}{2}-\sin^2\frac{x}{2} = 1-2\sin^2\frac{x}{2}$$ and $$\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$$ so $$\frac{1-\cos x}{\sin x} = \frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \tan\frac{x}{2}$$

• (provided $x\neq 0\mod \pi$, of course) – Clement C. Jul 10 '13 at 13:43
• When a problemis marked "homework" please don't answer the problem completely. – Thomas Andrews Jul 10 '13 at 13:44
• Oh. My apologies. – Clement C. Jul 10 '13 at 13:45
• Hi, im sorry to bug. Im still not understanding what you did on the first row with cosx=..... – Red Queen10101 Jul 10 '13 at 13:58
• You have $$\cos(u+v) = \cos u \cos v - \sin u\sin v$$ plug into this $u=v=\frac{x}{2}$. – Clement C. Jul 10 '13 at 14:01