Question

What is the answer of:-

√(2+(√2+√(2+2cosθ)))

Answer 1

Hey there!

Take a look at the formula of cos2θ.

cos2θ=2cos²θ-1

cos2θ+1=2cos²θ

NOW REMEMBER θ TAKES ANY VALUE

SO FOR LACK OF CONFUSION

PUT θ=x

cos2x+1=2cos²x


NOW BACK TO THE QUESTION


√(2+(√2+√(2+2cosθ)))

= √(2+(√2+√(2[1+cosθ] )))

Substituting cos2x+1=2cos²x we get

= √(2+(√2+√(2[2cos²θ/2 ] )))

=√(2+√(2+√4cos²θ/2))

=√(2+√(2+2cosθ/2))

=√(2+√(2(1+cosθ/2)))


=√(2+√(2(2cos²θ/4)))

=√(2+√(4cos²θ/4))

=√(2+2cosθ/4)

=√(2(1+cosθ/4))

=√(2(2cos²θ/8)

=√(4cos²θ/8)

=2cosθ/8

hope helped!

Answer 2

Hey there !!!!

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=√(2+(√2+√(2+2cosθ)))

cos2∅=2cos²∅-1

Similarly

cosθ=2cos²θ/2-1

1+cosθ=2cos²θ/2

=√(2+(√2+√(2+2cosθ)))

=√(2+(√2+√2(1+1cosθ)))

=√(2+(√2+√2*2cos²θ/2)))

=√(2+(√2+2cosθ/2))

1+cosθ/2=2cos²θ/4

=√(2+(√2(1+cosθ/2)))

=√(2+(√2(2cos²θ/4)))

=√(2+(√2(2cos²θ/4)))

=√(2+2cosθ/4)

=√(2+2cosθ/4)

1+cosθ/4=2cos²θ/8

=√2(1+cosθ/4)

=√2*2cos²θ/8=2cosθ/8

Note: Such problems for objective purpose can be solved with simple formula 
        
             √(2+(√2+√(2+2cosθ)))= 2cosθ/2ⁿ

Here n = number of terms square root is applied 

In  √(2+(√2+√(2+2cosθ))) n=3

 √(2+(√2+√(2+2cosθ)))=2cosθ/2ⁿ=2cosθ/2³=2cosθ/8

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 Hope this helped you.........