Check out some similar questions!

blondetrig · Nov 15, 2009, 09:15 AM

I need to prove the following

sinxcot(x/2)=cosx+1

I'm usually OK in math but have a huge trig block... help!

Nhatkiem · Nov 15, 2009, 12:58 PM

Please put parenthesis where needed. Is that cos(x) or cos (x+1), is that sin(x)cot(x/2) or sin(cot(x/2))?

Unknown008 · Nov 16, 2009, 12:06 AM

Well, what I would have done is:

1. Convert all into half angle, in terms of cos and sin only:

$sinxcot(\frac{x}{2}) = 2sin (\frac{x}{2}) cos (\frac{x}{2}) \frac{cos(\frac{x}{2})}{sin(\frac{x}{2})}$

Then, cross out the ones which cancel:

$2\cancel{sin (\frac{x}{2})} cos (\frac{x}{2}) \frac{cos(\frac{x}{2})}{\cancel{sin(\frac{x}{2})}} = 2cos^2(\frac{x}{2})$

Then, transform back to single angle using the identity:

$cos2A = 2cos^2 A -1$

This should lead you to the answer.

I hope it helped! :)

tiarav001 · Nov 19, 2009, 08:00 PM

how to solve 1+cot^2x/cscx=1/sinx

Unknown008 · Nov 19, 2009, 08:09 PM

What is $1+cot^2x$ ? That is the equivalent of $cosec^2x$
You have cosec^2x and cosecx which reduce to cosecx. What is the definition of cosec? That will give you the answer.