# Proving Trig Identities

• Feb 25th 2009, 08:41 AM
Proving Trig Identities
Hey Guys and Girls!
I need some help answering one of my homework questions. This is from Grade 12u Advanced Functions.
I've done other examples, but this one seems to confuse me.

Prove the identity:

sinxcosxtanx = 1 - cos^2x

This is what I've done so far:
Left Side:
=sinxcosx(sinx/cosx)

and from here I don't know what to do, but my guess would be:
= sin^2x / cos^2x

Which means both sides are not equal?

Right Side:
1 - cos^2x
=sin^2x

(Rofl)
• Feb 25th 2009, 09:26 AM
Amanda H
You've done it right but you've just made a small error. I'll right it out in full for you:
Left side
$\sin x \cos x \tan x = \sin x \cos x (\frac{\sin x}{\cos x})$
The above way of setting it out might allow you to see your error. Both $\cos x$ should cancel each other out

$= \sin x * \sin x$
$= \ sin^2 x$
$= 1 - \cos^2 x$
$= right side$
• Feb 25th 2009, 09:39 AM
Yes that is what I was thinking, but how come the cosx's cancel out yet the sinx's become sin^2x?
• Feb 25th 2009, 09:53 AM
Amanda H
You already have $\sin x \cos x$ and $\tan x = \frac{\sin x}{\cos x}$ so multiplying the two it becomes:

$\sin x \cos x * \frac{\sin x}{\cos x}$

$\frac{\sin x * \cos x * \sin x}{\cos x}$

leaving:

$\sin^2 x$

Just like using numbers:

$\frac{4}{5} * \frac{3}{4}$

$\frac{4*3}{5*4}$

here the 4 cancels leaving:

$\frac{3}{5}$

Does that make it clearer?
• Feb 25th 2009, 09:56 AM