Challenge Level

If $\theta+\phi+\psi=\tfrac{1}{2}\pi,$ show that $$ \sin^{2}\theta+\sin^{2}\phi+\sin^{2}\psi+2\sin\theta\sin\phi\sin\psi=1. $$

*Below are some hints about how you might approach this part of the question.*

It might help if you write down some of the trig identities that you know such as $\sin^2 \alpha + \cos^2 \alpha =1$, $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \sin \beta \cos \alpha$ etc.

Can you express $\cos \left(\frac{\pi} 2 - \alpha \right)$ in another way? It might help to sketch $\cos x$ and $\sin x$ in the range $0 \le x \le \frac{\pi} 2$. Can you apply this when $\alpha = \psi$?

What is the value of $\sin(\theta+ \phi + \psi)$?

There is more than one way to solve this problem, so you might not find all of the above helpful for your method, and you might need other identities as well!

Can you express $\cos \left(\frac{\pi} 2 - \alpha \right)$ in another way? It might help to sketch $\cos x$ and $\sin x$ in the range $0 \le x \le \frac{\pi} 2$. Can you apply this when $\alpha = \psi$?

What is the value of $\sin(\theta+ \phi + \psi)$?

There is more than one way to solve this problem, so you might not find all of the above helpful for your method, and you might need other identities as well!

By taking $\theta=\phi=\tfrac{1}{5}\pi$ in this equation, or otherwise, show that $\sin\tfrac{1}{10}\pi$ satisfies the equation $$ 8x^{3}+8x^{2}-1=0. $$ *Again, here are some hints.*

Note that $\theta = \phi = 2 \psi$.

The algebra is slightly easier if you start by using $[\sin 2 \psi]^2=[2 \sin \psi \cos \psi]^2 $.

You will probably end up with a quintic (polynomial of degree 5). This can be factorised so that one of the brackets is equal to the required cubic. You can use long division of polynomials if you want, but "inspection" is quicker and less error prone here!

The algebra is slightly easier if you start by using $[\sin 2 \psi]^2=[2 \sin \psi \cos \psi]^2 $.

You will probably end up with a quintic (polynomial of degree 5). This can be factorised so that one of the brackets is equal to the required cubic. You can use long division of polynomials if you want, but "inspection" is quicker and less error prone here!

*Based on STEP Mathematics I, 1991, Q1. Question reproduced by kind permission of Cambridge Assessment Group Archives. The question remains Copyright University of Cambridge Local Examinations Syndicate ("UCLES"), All rights reserved.*