Question

\[\left| \theta \right|<\Pi/16 , then \sqrt{2 +\sqrt{2+\sqrt{2+ 2 \cos 8 teta}}}\] is equal to

Answer 1

@satellite73 @apoorvk @ajprincess

Answer 2

@shivam_bhalla

Answer 3

the options are cos2theta 2costheta

Answer 4

@supercrazy92 will help u surely :)

Answer 5

lol,
@heena is going to help you

Answer 6

2+2cos8x
=2(1+cos8x)
=2(1+2cos^2(4x)-1)
=2*2c0s^2(4x)
=4cos^2(4x)
Find the root of it.

Answer 7

no no its a golden opportunity for u to get a medal from me go for it :P @supercrazy92

Answer 8

@ajprincess there are three 2 s ....

Answer 9

ya i knw that. did u find the root of it?

Answer 10

2cos2theta

Answer 11

No it is 2cos4theta
Nw do for 2+2cos4theta as I did for 2+2cos8theta.

Answer 12

2+2cos4theta
=2(1+cos4theta)
can u do nw?

Answer 13

it seems finished

Answer 14

i gave up!

Answer 15

show me yr work.

Answer 16

i dont ghave any idea can u tell wat u have done

Answer 17

2+2cos8x
=2(1+cos8x) (First I took 2 as the common factor)
=2(1+2cos^2(4x)-1) (using cos2x=2cos^2x-1 I wrote this)
=2(2cos^2(4x))
=4cos^2(4x)
Is it clear nw?

Answer 18

\[\large \cos \left( 2x \right)=\ \ \cos^2\left( x \right)-\sin^2\left( x \right)\]\[\large \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \cos^2\left( x \right)-\left( 1-\cos^2\left( x \right) \right)\]\[\LARGE \ \ \ \ \ \ \ \ \ \ =\ 2\cos^2\left( x \right)-1\]

keep using this rule, as ajprincess showed, to show what's under a square root as a perfect square: then simplify from the inside out