Question

Given cos2x=(-5/12) and pi/2 < x < pi, how do you find the values of the six trig functions?

Answer 1

See explanation for the list.

Explanation:

`x in Q2 and,` as `cos 2x=-5/12<0, 2x in Q3`.

In Q2, sine and its reciprocal cosecant (csc) are positive. Others are,

negative. So, given `cos 2x=-5/12`,

`sin x=sqrt((1-cos 2x)/2)=sqrt(17/24)`

`csc x=1 /sin x=sqrt(24/17)`

`cos x =-sqrt((1+cos 2x)/2)=-sqrt(7/24)`

`sec x = 1/cos x=-sqrt(24/7)`

`tan x=sin x/(cos x)=--sqrt(17/7)`

`cot x=1/(tan x)=-sqrt(7/17)`

In Q3, tangent and its reciprocal cotangent are positive. Others are

negative.

So, given `cos 2x=-5/12`,

`sec 2x =1/(cos 2x) =-12/5.`

`sin 2x= -sqrt(1-cos^2 2x)=-sqrt(1-25/144)=-sqrt119/12`

`csc 2x=1/(sin 2x)=-12/sqrt119`

`tan 2x=(sin 2x) /(cos 2x)=sqrt119/5`

`cot 2x=1/(tan 2x)=5/sqrt119`

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