
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
Thanks for any help you can give me!
(Rofl)

You've done it right but you've just made a small error. I'll right it out in full for you:
Left side
$\displaystyle \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 $\displaystyle \cos x $ should cancel each other out
$\displaystyle = \sin x * \sin x $
$\displaystyle = \ sin^2 x $
$\displaystyle = 1  \cos^2 x $
$\displaystyle = right side $

Yes that is what I was thinking, but how come the cosx's cancel out yet the sinx's become sin^2x?

You already have $\displaystyle \sin x \cos x $ and $\displaystyle \tan x = \frac{\sin x}{\cos x} $ so multiplying the two it becomes:
$\displaystyle \sin x \cos x * \frac{\sin x}{\cos x} $
$\displaystyle \frac{\sin x * \cos x * \sin x}{\cos x} $
leaving:
$\displaystyle \sin^2 x $
Just like using numbers:
$\displaystyle \frac{4}{5} * \frac{3}{4} $
$\displaystyle \frac{4*3}{5*4} $
here the 4 cancels leaving:
$\displaystyle \frac{3}{5} $
Does that make it clearer?

Oh I get it....Thank you so much (Rock)