jonnburton
Junior Member
- Joined
- Dec 16, 2012
- Messages
- 155
Hi
I'm having more trouble with trig, in particular equations.
Would anyone be able to point me in the right direction with the following:
given that \(\displaystyle 2cos^2x - sin^2x = 1\), show that \(\displaystyle cos^2x =2 sin^2x\) and hence find the possible values of cot x
I did the first part:
\(\displaystyle 2cos^2x - sin^2x = cos^2 x + sin^2x\)
\(\displaystyle cos^2x -2sin^2x = 0\)
\(\displaystyle cos^2x = 2sin^2x\)
But I stuck on working from this to getting any values for cot x.
I'd be very grateful for any help!
I'm having more trouble with trig, in particular equations.
Would anyone be able to point me in the right direction with the following:
given that \(\displaystyle 2cos^2x - sin^2x = 1\), show that \(\displaystyle cos^2x =2 sin^2x\) and hence find the possible values of cot x
I did the first part:
\(\displaystyle 2cos^2x - sin^2x = cos^2 x + sin^2x\)
\(\displaystyle cos^2x -2sin^2x = 0\)
\(\displaystyle cos^2x = 2sin^2x\)
But I stuck on working from this to getting any values for cot x.
I'd be very grateful for any help!
