Verifying Trigonometric Identities

• May 4th 2009, 08:28 PM
AlderDragon
Verifying Trigonometric Identities
I'm a bit confused on verifying trigonometric identities. An example problem would be:

$\cos(\frac{\pi}{2})-\Theta=\sin\Theta$

Or:

$\cot(\pi+\Theta)=\cot\Theta$

Can someone please explain what exactly the question is asking, and how I find the solution? Thanks.

Also, is there any way to visually differ between the graph of a sine function and the graph of a cosine function, if you don't know if one may be shifted? Sorry if that is hard to understand.
• May 4th 2009, 10:44 PM
vemrygh
The question is asking to show that the expressions of LHS and RHS are equal, i.e. that you can use some identities to show equality.
Take the first one:

$\cos(\frac{\pi}{2}-\theta)=
\cos(\frac{\pi}{2})\cos\theta+\sin(\frac{\pi}{2})\ sin\theta
=\sin\theta$
.

For the graph question. I distinct between the sine and cosine graphs by remembering that sine passes through the origin, while cosine does not pass through the origin.
• May 5th 2009, 05:13 AM
AlderDragon
Quote:

Originally Posted by vemrygh
The question is asking to show that the expressions of LHS and RHS are equal, i.e. that you can use some identities to show equality.
Take the first one:

$\cos(\frac{\pi}{2}-\theta)=
\cos(\frac{\pi}{2})\cos\theta+\sin(\frac{\pi}{2})\ sin\theta
=\sin\theta$
.

For the graph question. I distinct between the sine and cosine graphs by remembering that sine passes through the origin, while cosine does not pass through the origin.

I still don't understand how to solve those problems. For:
$

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

How do you get from the original problem to:

$
\cos(\frac{\pi}{2}-\theta)\cos\Theta
$

And from that to:

$
\cos(\frac{\pi}{2})\cos\theta+\sin(\frac{\pi}{2})\ sin\theta
$
• May 5th 2009, 08:24 AM
running-gag
Quote:

Originally Posted by AlderDragon

I still don't understand how to solve those problems. For:
$

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

How do you get from the original problem to:

$
\cos(\frac{\pi}{2}-\theta)\cos\Theta
$

And from that to:

$
\cos(\frac{\pi}{2})\cos\theta+\sin(\frac{\pi}{2})\ sin\theta
$

Hi

vemrygh used the trig identity cos(a-b)=cos a cos b + sin a sin b
with a = pi/2 and b = theta