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

$sin 16 x = - \frac{4}{5}, and, cos 4 x = \frac{1}{2} .$ $sin16x=-4/5, and, cos4x=1/2.$

Given that, $cot 8 x = - 2 \Rightarrow tan 8 x = \frac{1}{cot 8 x} = - \frac{1}{2} .$ $cot8x=-2 rArr tan8x=1/(cot8x)=-1/2.$

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

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

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