# Proving Trigo Identity

• November 28th 2009, 10:22 PM
gp3
Proving Trigo Identity
Hi guys,
how do you prove these 2:

http://img29.imageshack.us/img29/5765/scan0001ik.jpg
• November 28th 2009, 10:37 PM
alexmahone
Quote:

Originally Posted by gp3
Hi guys,
how do you prove these 2:

http://img29.imageshack.us/img29/5765/scan0001ik.jpg

LHS

= $\frac{sin3x}{sinx}+\frac{cos3x}{cosx}$

= $\frac{sin3xcosx+cos3xsinx}{sinxcosx}$

= $\frac{sin(3x+x)}{sinxcosx}$

= $\frac{sin4x}{sinxcosx}$

= $\frac{2sin2*2x}{2sinxcosx}$

= $\frac{2*2sin2xcos2x}{sin2x}$

= $4cos2x$

= RHS

-----------------------------------------------------------------------------------

RHS

= $\sqrt{\frac{1}{2}(1-sin\theta)}$

= $\sqrt{\frac{1}{2}(cos^2\frac{\theta}{2}+sin^2\frac {\theta}{2}-2sin\frac{\theta}{2}cos\frac{\theta}{2})}$

= $\sqrt{\frac{1}{2}(cos\frac{\theta}{2}-sin\frac{\theta}{2})^2}$

= $\frac{1}{\sqrt2}cos\frac{\theta}{2}-\frac{1}{\sqrt2}sin\frac{\theta}{2}$

= $sin\frac{\pi}{4}cos\frac{\theta}{2}-cos\frac{\pi}{4}sin\frac{\theta}{2}$

= $sin(\frac{\pi}{4}-\frac{\theta}{2})$

= LHS
• November 28th 2009, 10:53 PM
Hello gp3
Quote:

Originally Posted by gp3
Hi guys,
how do you prove these 2:

http://img29.imageshack.us/img29/5765/scan0001ik.jpg

$\frac{\sin3x}{\sin x}+ \frac{\cos3x}{\cos x}$
$=\frac{\sin3x\cos x +\cos 3x \sin x}{\sin x \cos x}$

$=\frac{\sin(3x+x)}{\tfrac12\sin2x}$

$=\frac{2\sin4x}{\sin2x}$

$=\frac{4\sin2x\cos2x}{\sin2x}$

$=4\cos2x$
$\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)$
$=\sin\frac{\pi}{4}\cos\frac{\theta}{2}-\cos\frac{\pi}{4}\sin\frac{\theta}{2}$

$=\frac{1}{\sqrt2}\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right)$

$=\sqrt{\frac{1}{2}\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right)^2}$

$=\sqrt{\frac{1}{2}\left(\cos^2\frac{\theta}{2}-2\sin\frac{\theta}{2}\cos\frac{\theta}{2}+\sin^2\f rac{\theta}{2}\right)}$

$=\sqrt{\frac{1}{2}\left(1-\sin\theta\right)}$
• November 28th 2009, 11:21 PM
gp3
hi guys, thanks for your help.

I dont understand how you got this part: =\frac{\sin(3x+x)}{\tfrac12\sin2x}

and this part: =\frac{1}{\sqrt2}\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right
• November 28th 2009, 11:39 PM
Hello gp3
Quote:

Originally Posted by gp3
hi guys, thanks for your help.

I dont understand how you got this part: =\frac{\sin(3x+x)}{\tfrac12\sin2x}

$\sin (A+B) = \sin A \cos B + \cos A \sin B$

Quote:

and this part: =\frac{1}{\sqrt2}\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right
$\sin \pi/4 = \cos \pi/4 = 1/\sqrt2$

• November 29th 2009, 12:24 AM
gp3
http://www.mathhelpforum.com/math-he...ccb5afa9-1.gif
erm, i don't understand why is the part in bracket squared?

=\sqrt{\frac{1}{2}\left(1-\sin\theta\right)}
and how did you get to 1-Sin?

sorry for so many questions, i'm rather weak in my maths!
• November 29th 2009, 12:58 AM
Hello gp3
Quote:

Originally Posted by gp3
http://www.mathhelpforum.com/math-he...ccb5afa9-1.gif
erm, i don't understand why is the part in bracket squared?

$=\frac{1}{\sqrt2}\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right)$

$=\frac{1}{\sqrt2}\color{red}\sqrt{\left(\color{bla ck}\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\color{red}\right)^2}$

$=\sqrt{\frac{1}{2}\left(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\right)^2}$

Quote:

=\sqrt{\frac{1}{2}\left(1-\sin\theta\right)}
and how did you get to 1-Sin?

sorry for so many questions, i'm rather weak in my maths!
$\cos^2\frac{\theta}{2} + \sin^2\frac{\theta}{2} = 1$

Keep at it - it will come in time!