Verified answer

cos(A+B) = cosAcosB - sinAsinB

cos(A-B) = cosAcosB + sinAsinB

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cos(A+B) + cos(A-B) = 2cosAcosB

A + B = 6θ

A - B = 4θ

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A = 5θ

B = θ

cos(4θ) +cos(6θ) = 2cos(5θ)cos(θ)

2cos(5θ)cos(θ) = 0

cos(5θ)cos(θ) = 0

cos(5θ) = 0

5θ = {π/2, 3π/2, 5π/2, 7π/2, 9π/2, 11π/2, 13π/2, 15π/2, 17π/2, 19π/2}

θ = {π/10, 3π/10, 5π/10, 7π/10, 9π/10, 11π/10, 13π/10, 15π/10, 17π/10, 19π/10}

θ = {π/10, 3π/10, π/2, 7π/10, 9π/10, 11π/10, 13π/10, 3π/2, 17π/10, 19π/10}

OR

cos(θ) = 0

This has solutions of:

θ = {π/2, 3π/2}

θ = {π/10, 3π/10, π/2, 7π/10, 9π/10, 11π/10, 13π/10, 3π/2, 17π/10, 19π/10}

The answer is no. Plug in pi and we get cos(4pi) = 1 and cos(6pi) = 1. So 1 + 1 = 2 this does not equal 0.

A sad choice of login... I do hope the world's not been that cruel to you :-)

To make things more compact, let's write x=theta? then you have cos(4x)+cos(6x)=0 for any x E (0, 2pi).

First let's see if there are any solutions... a quick way is to make a plot of the function y= cos(4x) + cos(6x) over the interval (0, 2pi).... I get zero crossings near

x/pi E {0.1, 0.3, 0.7, 0.9, 1.1, 1.3, 1.5, 1.7, 1.9}... which means there are a whole lot of solutions!

There are trig identities for sums of angles... here's the trick

cos(4x)= cos(5x-x) = cos(5x)cos(x) + sin(5x)sin(x) and...

cos(6x) =cos(5x+x) = cos(5x)cos(x) - sin(5x)sin(x)

so that cos(4x)+cos(6x) == 2 cos(5x)cos(x). !!!

Ok, so now you have your easy to solve equation - if either cos(5x)=0 or cos(x)=0 you get a zero.

cos(5x) -> 0 whenever 5x = +/- pi.. and

cos(x) -> 0 whenever x = +/- pi

hope this helps. Tell your teacher this was a stinker!

noisejammer