Question

Show that cos²π/10+cos²4π/10+cos² 6π/10+cos²9π/10=2. I am a bit confused if I make Cos²4π/10=cos²(π-6π/10) & cos²9π/10=cos²(π-π/10), it will turn negative as cos(180°-theta)=-costheta in the second quadrant. How do I go about proving the question?

Answer 1

Please see below.

Explanation:

`LHS=cos^2(pi/10)+cos^2((4pi)/10)+cos^2((6pi)/10)+cos^2((9pi)/10)`

`=cos^2(pi/10)+cos^2((4pi)/10)+cos^2(pi-(4pi)/10)+cos^2(pi-(pi)/10)`

`=cos^2(pi/10)+cos^2((4pi)/10)+cos^2(pi/10)+cos^2((4pi)/10)`

`=2*[cos^2(pi/10)+cos^2((4pi)/10)]`

`=2*[cos^2(pi/2-(4pi)/10)+cos^2((4pi)/10)]`

`=2*[sin^2((4pi)/10)+cos^2((4pi)/10)]`

`=2*1=2=RHS`

Answer 2

We know that,
`color(red)(costheta = sin (pi/2-theta)` so also,
`color(red)(cos^2theta = sin^2 (pi/2-theta)`
`color(magenta)(costheta = -sin((3pi)/2-theta)` so also,
`color(magenta)(cos^2theta = (-sin((3pi)/2-theta))^2 = sin^2((3pi)/2-theta)`

getting back to the question,

`color(red)(cos²π/10)+cos²(4π)/10+cos² (6π)/10+color(magenta)(cos²(9π)/10)=2`

`color(red)(sin²(pi/2-π/10))+cos²(4π)/10+cos² (6π)/10+color(magenta)((-sin((3pi)/2-(9π)/10))^2)=2`

`sin²((5pi)/10-π/10)+cos²(4π)/10+cos² (6π)/10+sin²((3pi)/2-(9π)/10)=2`

`[sin²(4π)/10+cos²(4π)/10]+[cos² (6π)/10+sin²((15pi)/10-(9π)/10)]=2`

`[sin²(4π)/10+cos²(4π)/10]+[cos² (6π)/10+sin²(6π)/10]=2`

Applying, `sin^2theta + cos^2theta = 1`

`1+1=2`
`2=2`

Hence Proved.

P.S. you were going right, just note that even if its negative, the final answer turns out to be positive as the `cos` is squared according to the question. Any negative number squared is positive :)