Stumped on this Proof problem

[airforceone]

Hello,
I'm completely stumped on this proof problem:

The only step I can think of is:

Which is an Double Angle Identity.
I can't think of anything else from there.

Thanks!

 

[o_O]

Try starting on the right side. Usually, converting into terms of sinx and cosx would result in some cancelling.

Also, what identity involves \(\displaystyle 1 + tan^{2}x\)?

 

[airforceone]

Ok so I worked out the right side and got this:

So how would I convert it so that I worked on the left side?

Thanks!

 

[o_O]

Are you suppose to work starting from the left side? Well you could always "cheat" and copy what you did on the right side and work backwards :lol:

\(\displaystyle sin2x\)

\(\displaystyle = 2sinxcosx\)

\(\displaystyle =\frac{2sinx}{cosx} \cdot cos^{2}x\)
etc. etc.

But I don't see why you would need to start on the left side. If this is an identity, which we proven that it is, then it shouldn't matter which side you start from at all!

 

[airforceone]

Ya I'm supposed to work on the left side lol but theres not really a way to work on the left side without working the right side right?
She might get suspicious why I happened to just multiply cos^2x lol... :lol:

Thanks!

 

[stapel]

She might get suspicious why I happened to just multiply cos^2x

True enough. But haven't there been proofs in your book or at the board, where you were left wondering, "How on earth did they know to do that??" :shock:

You'd just be doing the same thing: Working both sides downward until they "match", and then, on a clean sheet of paper, copying down what you did for one side, jumping across the "equals" sign to the other side where they'd "matched", and then copying up the other side, until you get to that side's starting point. :wink:

Eliz.

 

[airforceone]

Ok thanks guys!