Question

Prove , cos20cos40cos80=1/8 . Angles are in degrees .

Answer 1

by talk about "trickonometry" who thought up this question?

Answer 2

one gimmick is to call
\[x=\cos(20)\cos(40)\cos(80)\] then write
\[\sin(20)x=\sin(20)\cos(20)\cos(40)\cos(80)\]

Answer 3

now use the "double angle" formula for sine twice. that is use
\[\sin(2x)=2\sin(x)\cos(x)\] so you get
\[\sin(20)x=\frac{1}{2}\sin(40)\cos(40)\cos(80)\] then again and get
\[\sin(20)x=\frac{1}{4}\sin(80)\cos(80)\] a
nd and again to get
\[\sin(20)x=\frac{1}{8}\sin(160)\]
finally a miracle occurs, namely that
\[\sin(20)=\sin(160)\] since 20+160=180
giving you
\[\sin(20)x=\frac{1}{8}\sin(20)\] cancel to get
\[x=\frac{1}{8}\]

Answer 4

Thank you !

Answer 5

yw

Answer 6

But what about sin10sin50sin70 ?

Answer 7

Figured it out , thank you again !