Submitted by helpmeh on Wed, 01/11/2017 - 17:57

I need help with STEP 1 2003 - Question 3 iii.

I completed the square and found that cos(theta/2) = 1/2 - root3/2.

I have reached the stage where I have that half theta = arcos(1/2-root3/2)

From here I'm not sure how to reach end conclusion.

Thank you.

## Almost there!

You have one value of $\frac 1 2 \theta$, i.e.

\[

\frac 1 2 \theta = \arccos \left( \frac 1 2 - \frac {\sqrt{3}} 2\right)

\]

However, $\cos x=k$ has infinitely many solutions (draw a sketch!), and if $x=\alpha$ is one

then so is $x=-\alpha $ and $x=\alpha+2\pi$ etc. (look at your sketch!) The general solutions here are $x=\alpha + 2n\pi$ and $x=-\alpha + 2n\pi$. (Look at your sketch to convince yourself!)

Relating this back to your problem, you have the general solutions:

\[

\frac 1 2 \theta = \arccos \left( \frac 1 2 - \frac {\sqrt{3}} 2\right) + 2n\pi

\]

and

\[

\frac 1 2 \theta = -\arccos \left( \frac 1 2 - \frac {\sqrt{3}} 2\right) + 2n\pi

\]

This is starting to look a little better - if we multiply by 2 we will get a $4n\pi$ which is promising.

The next bit is to try to relate $\arccos \left( \frac 1 2 - \frac {\sqrt{3}} 2\right)$ to $\phi$. Have a play and see if you can get this bit out.

## Thank you very much, your

Thank you very much, your response was very helpful.