Suppose θ is an angle strictly between 0 and π/2 such that sin5θ=(sin^5)θ. The number tan2θ can be uniquely written as a√b, where a and b are positive integers, and b is not divisible by the square of a prime. What is the value of a+b?

Answer 1

well, let's see. Leaving out all the x's,

sin(5x) = sin^5 + 5sin cos^4 - 10 sin^3 cos^2
= sin^5 + 5sin^5 - 10sin^3 + 5sin - 10sin^3 + 10sin^5
= 16sin^5 - 20sin^3 + 5sin

so, if sin(5x) = sin^5(x)
3sin^5 - 4sin^3 + sin = 0
(3sin-1)(sin-1) = 0
sin(x) = 1/3 or 0

x=0 is out, since it's not in (0,π/2)
So, sin(x) = 1/3
cos(x) = √8/3

sin(2x) = 2 sinx cosx = 2√8/9
cos(2x) = 1 - 2sin^2(x) = 7/9

tan(2x) = 2√8/7

Hmmm. I don't get a√b
Better double check my algebra.

Answer 2

the answer is 2*sqrt(3) or 3*sqrt(2) or 2*sqrt(2) or 3*sqrt(3).....try them.....

Answer 3

i still dun get it :(

Answer 4

copying from above
sin(5x) = = 16sin^5 - 20sin^3 + 5sin
sin(5x) = sin^5(x)
3sin^5 - 4sin^3 + sin = 0
(3sin^2 -1)(sin^2 -1) = 0
sin^2(x) = 1/3
tan x = 1/√2
tan 2x = 2√2