Double Angle Trig: Find Values of sin2x, cos2x, tan2x -90<x<180

[brandon1]

Homework Statement

(x=theta for typing here)
Find exact values of sin2x, cos2x, and tan2x when sec(x)=-6; 90degrees<x<180degrees

Homework Equations

sin(2x)=2sinxcosx
etc..

The Attempt at a Solution

If I can figure out what I am doing wrong for just sin(x) I should be good to go.

sec(x)=-6 therefore cos(x)=-1/6

sin(2x)=(2)((sqr37)/36)(-1/6)

sin2x=(-2sqr37)/36

I have done this problem EXACTLY how one like it is done in the notes, and keep coming up with the incorrect answer. To check myself sin2x=-.338 and what I got for sin2x=-.328

 

[cristo]

where does sqrt(37/36) come from? I presume you used sin²x+cos²x=1, and used cosx=-1/6. This would give sin²x=1-(-1/6)²=35/36.

 

[brandon1]

where does sqrt(37/36) come from? I presume you used sin²x+cos²x=1, and used cosx=-1/6. This would give sin²x=1-(-1/6)²=35/36.

I used the pythagereon therom.

a^2+b^2=c^2
b^2=c^2-a^2
b^2=6^2--1^2
b=sqr37

wouldn't I have to take the square root of 35/36 using pythag id to get sin(x)?


Edit:Ok, I got the correct answer, but I don't undestand why the pythag therom did not work...

 

[jing]

I used the pythagereon therom.

a^2+b^2=c^2
b^2=c^2-a^2

what is the value of a^2 when a = -1?

Remember you square first then subtract!

 

[brandon1]

what is the value of a^2 when a = -1?

Remember you square first then subtract!

Careless mistakes can be lethal