1. ## C3 Trigonometry Question

Here is a question, I've got an answer to it but I think its wrong

a) prove that (done)

$\frac{1-cos2\Theta }{sin2\Theta }\equiv tan\Theta$

b) verify that

$\Theta =180$ is a solution to the equation

$sin2\Theta= 2-2cos2\Theta$

Does this rearrange to, using part (a), $2tan\Theta =0$ so, $2tan180 =0$, which is correct??

c) Using your result in part (a), or otherwise, find the other two solutions, $0< \Theta < 360$, of the equation using $sin2\Theta= 2-2cos2\Theta$

For this I said that this meant that $tan\Theta =0$ and the inverse tan of 0 = 0

So, using the general solution for tan, $\Theta = 180n + \beta$ and the fact that $\beta = 0$

I got the answers 0 and 360, but I'm doubtful as the range is $0< \Theta < 360$ with no 'greater/less than or equal to' in there??

2. ## Re: C3 Trigonometry Question

Hello there, it is sufficient for part b) just to check that the LHS and the RHS agree. So considered separately, when theta=180 , the LHS sin(360)=0 and the RHS 2-2cos(360) = 0. This shows that theta=180 is a solution; therefore, there is no need for arrangement.

For part c), I think you have a made a small mistake when rearranging - it should be 1= (2-2 cos(2 theta)) / ( sin 2 theta) which gives tan theta = 0.5.