Help with this trigonometry problem

Solution 1:

A method using trigonometric identities, although a very long one -

For simplicity, let me put $\dfrac{\pi}{70}=A$. Now, the expression is -

$(\sin{9A}+\sin{29A}-\sin{31A})(\sin{A}-\sin{11A}-\sin{19A})$

Multiply out both the brackets completely and for each term of the type "$\sin*\sin$", convert it into sum of two cosines using the formula -

$2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}$, to get the following expression -

$\dfrac{1}{2}(\cos{8A}-\cos{10A}-\cos{2A}+\cos{20A}-\cos{10A}+\cos{28A}+\cos{28A}-\cos{30A}-\cos{18A}+\cos{40A}-\cos{10A}+\cos{48A}-\cos{30A}+\cos{32A}+\cos{20A}-\cos{42A}+\cos{12A}-\cos{50A})$

Now, convert all those cosines whose angles are greater than 90 degrees (i.e. greater than $35A$ here), into cosines of angles less than a 90 degrees (i.e. less than $35A$ here), by using the fact that $\cos{(180-A)}=-\cos{A}$, to get the following simplified expression -

$\dfrac{1}{2}(\cos{8A}-3\cos{10A}-\cos{2A}+3\cos{20A}+3\cos{28A}-3\cos{30A}-\cos{18A}-\cos{22A}+\cos{32A}+\cos{12A})$, which further becomes -

$\dfrac{1}{2}((\cos{8A}-\cos{22A})-(\cos{2A}+\cos{18A})+(\cos{32A}+\cos{12A})-3\cos{10A}+3\cos{20A}-3\cos{30A}+3\cos{28A})$

$\dfrac{1}{2}((\cos{8A}-\cos{22A})-2\cos{10A}\cos{8A}+2\cos{22A}\cos{10A}-3\cos{10A}+3\cos{20A}-3\cos{30A}+3\cos{28A})$

$\dfrac{1}{2}((\cos{8A}-\cos{22A})(1-2\cos{10A})-3\cos{10A}+3\cos{20A}-3\cos{30A}+3\cos{28A})$

$\dfrac{1}{2}((2\sin{15A}\sin{7A})(1-2\cos{10A})-3\cos{10A}+3\cos{20A}-3\cos{30A}+3\cos{28A})$

$\dfrac{1}{2}((2\sin{7A})(\sin{15A}-2\sin{15A}\cos{10A})-3\cos{10A}+3\cos{20A}-3\cos{30A}+3\cos{28A})$

$\dfrac{1}{2}((2\sin{7A})(\sin{15A}-\sin{25A}-\sin{5A})-3\cos{10A}+3\cos{20A}-3\cos{30A}+3\cos{28A})$

Now use: $\sin{15A}=\cos{20A},\sin{25A}=\cos{10A},\sin{5A}=\cos{30A}$ (Complementary angle pairs, check for yourself!) to write the above expression as -

$\dfrac{1}{2}((2\sin{7A})(\cos{20A}-\cos{10A}-\cos{30A})+3(\cos{20A}-\cos{10A}-\cos{30A})+3\cos{28A})$

$\dfrac{1}{2}((2\sin{7A}+3)(\cos{20A}-\cos{10A}-\cos{30A})+3\cos{28A})$

Now use: $\cos{20A}=-\cos{50A}$ (supplementary angle pairs, check for yourself!) and rewrite the above expression as -

$\dfrac{1}{2}((2\sin{7A}+3)(-\cos{10A}-\cos{30A}-\cos{50A})+3\cos{28A})$ ... (1)

$\cos{10A}+\cos{30A}+\cos{50A}=\dfrac{\cos{30A}\times\sin{30A}}{\sin{10A}}=\dfrac{\sin{60A}}{2\sin{10A}}=\dfrac{1}{2}$. Put this in expression (1) to get -

$\dfrac{1}{2}(-\dfrac{1}{2}(2\sin{7A}+3)+3\cos{28A})$

But $\cos{28A}=\sin{7A}=\dfrac{\sqrt{5}-1}{4}$. Finally put these to get your answer!