How do you find sin(16x) and cos(4x), if you know that cot(8x) = -2 and 8x is an element of II? Thank you in advance!

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Answer 1 · 2017-11-16T18:20:17

# sin16x=-4/5, and, cos4x=1/2.#

Given that, #cot8x=-2 rArr tan8x=1/(cot8x)=-1/2.#

Recall that, # sin2theta=(2tantheta)/(1+tan^2theta).........(star^1),#

# cos2theta=(1-tan^2theta)/(1+tan^2theta)..........(star^2),#

# tan2theta=(2tantheta)/(1-tan^2 theta)..............(star^3).#

With #theta=8x" in "(star^1),# we have,

#sin(2(8x))=sin16x=(2tan8x)/{1+tan^2 8x)=(2(-1/2))/(1+1/4)=-4/5.#

Sub.ing, #theta=4x" in "(star^3),# we get,

#tan(2(4x))=tan 8x=(1-tan^2 4x)/(1+tan^2 4x)." But, "tan8x=-1/2.#

#:. -1/2=(1-tan^2 4x)/(1+tan^2 4x).#

#:. 2tan^2 4x-2=tan^2 4x+1, or, tan^2 4x=3.#

# :. tan 4x=+-sqrt3.#

But, # pi/2 lt 8x lt pi rArr pi/4 lt 4x lt pi/2. :. tan4x=+sqrt3.#

#:. sec^2 4x=1+tan^2 4x=4 ;. sec4x=+-2 :. cos4x=+-1/2.#

#"But, "pi/4 lt 4x lt pi/2. :. cos4x=+1/2.#