# Trigonometry - half angles problems

1. Oct 14, 2011

### paul18

1. The problem statement, all variables and given/known data
Previous part of the question I have solved:
Express cos(2x) in terms of sin(x):

cos(2x)=1-2sin2(x)

Hence or otherwise solve the equation
cos(x) + 3sin(x/2) = 2

2. Relevant equations
Double angle formulae:
cos(2x)=cos(2x) - sin(2x)
sin(2x)=2sin(x)cos(x)

3. The attempt at a solution
So basically I'm doing revision on trig, and I know I've come across this problem before, where you are presented with a full angle and a half angle, but I have failed to find an example. It's something to do with halving a double angle formula I think but I can't even start it. How do I get rid of the half angle? I know I should show an attempt, but I can't even start it off. Just a refresher on how to remove half angles would be great thanks :)
(Yes i have read the rules but this problem is difficult to put into that format properly)

2. Oct 14, 2011

### sjb-2812

If the first part read cos(2z) in terms of sin(z), what would your answer be? Then substitute z for x/2 in the second question

3. Oct 14, 2011

### eumyang

Note that x = 2 times (x/2). It may be helpful if you substitute another variable, say y, in for x/2. What would x equal? What does this equation:
cos(x) + 3sin(x/2) = 2
look like with the substitutions?

4. Oct 14, 2011

### paul18

cos(2y) + 3sin(y) = 2.
Thanks very much that's what I needed a refresher on :)