# trignometric identities

• Nov 23rd 2011, 08:34 PM
earthboy
trignometric identities
prove that:
1. $\frac{\sin A + \sin 2A}{\cos A - \cos 2A} =\cot (A/2)$
2. $\frac{1+\cos A}{\sin A} = \cot (A/2)$

• Nov 23rd 2011, 08:47 PM
pickslides
Re: trignometric identities
What have you tried to this point?

I suggest thinking about some double angle formulas for the first question.
• Nov 23rd 2011, 09:14 PM
earthboy
Re: trignometric identities
i tried using the double angle formulas but dont know how to bring about the cot(A/2) for the first one .
• Nov 24th 2011, 12:03 AM
Siron
Re: trignometric identities
For the first one I think it can be useful to work with the identities:
$\sin(x)+\sin(y)=2\sin\left(\frac{x+y}{2}\right)$ $\cos\left(\frac{x-y}{2}\right)$
$\cos(x)-\cos(y)=-2\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)$
• Nov 24th 2011, 03:59 AM
earthboy
Re: trignometric identities
i did the fiirst one.
• Nov 24th 2011, 05:42 AM
e^(i*pi)
Re: trignometric identities
Quote:

Originally Posted by earthboy
prove that:

2. $\frac{1+\cos A}{\sin A} = \cot (A/2)$

Use your double angle identities for sin(A) and cos(A) noting that $\cos(A) = \cos \left(2 \cdot \dfrac{A}{2}\right)$

Since you want cot then you want to get cos(A/2) in the numerator rather than sin(A/2)
• Nov 24th 2011, 06:56 AM
Soroban
Re: trignometric identities
Hello, earthboy!

Quote:

$\text{Prove that: }\:\frac{1+\cos A}{\sin A} \:=\: \cot \frac{A}{2}$

Right side: . $\cot\frac{A}{2} \;=\;\frac{\cos\frac{A}{2}}{\sin\frac{A}{2}} \;=\;\frac{\sqrt{\frac{1+\cos A}{2}}}{\sqrt{\frac{1-\cos A}{2}}} \;=\;\sqrt{\frac{1+\cos A}{1-\cos A}}$

Multiply by $\tfrac{1+\cos A}{1+\cos A}\!:\;\;\sqrt{\frac{1+\cos A}{1-\cos A}\cdot\frac{1+\cos A}{1+\cos A}} \;=\;\sqrt{\frac{(1+\cos A)^2}{1-\cos^2A}}$

. . . . . . . . . . . . $=\;\sqrt{\frac{(1+\cos A)^2}{\sin^2\!A}} \;=\; \frac{1+\cos A}{\sin A}$