Question

help in trig identities please!
cos4xcosx-sin4xsinx=0
My calculus friend came up with the answer of
(pi/2)-x=4x, resulting in x=pi/10. I graphed it, and it was an answer within the interval of [0,2pi] But he didn't explain his work on how he arrived at that answer. Can someone help please?

Answer 1

Do you know of a trig identity that has this form?
__ = cos a cos b - sin a sin b

Answer 2

nope, which identity is that?

Answer 3

When I was trying to solve this, I tried using the double angle formulas, but I couldn't get it to work.

Answer 4

\[\cos \left( a+b \right)=\cos a \cos b-\sin a \sin b\]

Answer 5

so would that be applying it to \[\cos 4x=\cos2\cos2-\sin2\sin2\]?

Answer 6

\[\cos 4x \cos x-\sin 4x \sin x=\cos \left( 4x+x \right)=\cos 5x=0\]

Answer 7

Oh ok. I see how thats applied now. Thanks, it makes the problem alot more solvable once sin is out.

Answer 8

\[\cos 5x=0=\cos \left(2 n+1 \right)\frac{ \pi }{2 },5x=\left( 2n+1 \right)\frac{ \pi }{ 2 }\]
\[x=\left( 2n+1 \right)\frac{ \pi }{ 10 }\],n=0,1,2,...,9

Answer 9

Ok, I got it know thanks alot for your help.

Answer 10

yw