Prove that $\cos24^\circ\cos84^\circ=\cos^2 72^\circ $

Question

I simplified this to:

$\sqrt {150+30\sqrt 5} + 14 + 9\sqrt{6-2\sqrt5}= 10\sqrt5+\sqrt{30+6\sqrt5} + 2\sqrt{30-6\sqrt5}+\sqrt{30-10 \sqrt5}$

any thoughts

Answer 1

We need to prove that $$\cos108^{\circ}+\cos60^{\circ}=1+\cos144^{\circ}$$ or $$\cos108^{\circ}-\cos144^{\circ}=\frac{1}{2}$$ or $$2\sin18^{\circ}\sin126^{\circ}=\frac{1}{4}$$ or $$\sin18^{\circ}\cos36^{\circ}=\frac{1}{4}$$ or $$\frac{\sqrt5-1}{4}\cdot\frac{\sqrt5+1}{4}=\frac{1}{4},$$ which is obvious.

Id est, your statement is true.

Done!

About your identity.

We need to prove that $$\sqrt {150+30\sqrt 5} + 14 + 9(\sqrt5-1)= 10\sqrt5+\sqrt{30+6\sqrt5} + 2\sqrt{30-6\sqrt5}+\sqrt5(\sqrt5-1)$$ or $$\sqrt {150+30\sqrt 5}=\sqrt{30+6\sqrt5} + 2\sqrt{30-6\sqrt5}$$ or $$150+30\sqrt5=30+6\sqrt5+4\sqrt{(30+6\sqrt5)(30-6\sqrt5)}+4(30-6\sqrt5)$$ or $$48\sqrt5=4\sqrt{900-180}.$$ Done!

Answer 2

Hint: use the sum to product formula for the left hand side...

$$\cos \alpha \cos \beta = \frac {cos (\alpha - \beta) + \cos (\alpha + \beta)}{2}$$

Then use the power reduction formula for cosine on the right hand side...

$$\cos^2 \alpha = \frac {1 + \cos 2 \alpha}{2}$$