Question

How to prove that

(sin9π70+sin29π70−sin31π70)(sinπ70−sin11π70−sin19π70)=5–√−44?


helpppp​

Answer 1

Here is an answer, though I'll be amazed if no-one can provide something more elegant. Let α=eiπ/70. Then α70=−1 and we have

LHS=(α9−α−92i+etc)(etc) ,

so

−4(LHS)=(α9+α61+etc)(etc) .

Multiplying everything out on the RHS gives the sum of all αk, where k belongs to the following sets:

{2,22,42,62,82,102,122}{18,38,58,78,98,118,138}{42,98}twice{10,30,50,90,110,130}three times.

Now the numbers αk for the first set are seven points equally spaced around the unit circle, so the sum is zero. Same goes for the second set. Same for the fourth, except that 70 is missing - no it wasn't a typo! So the sum of αk for the fourth set is −α70, that is, 1. So we have

−4(LHS)=2(α42+α98)+3=4cos3π5+3 ,

and since it is known that 4cos(3π/5)=1−5–√, we are done.

$\red{hope \: this \: helps}$

Answer 2

Answer:

Refer to the Attachment.

Hope it Helps !!!!