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Double Angle Trig

  • Thread starter brandon1
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  • #1
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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
 

Answers and Replies

  • #2
cristo
Staff Emeritus
Science Advisor
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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.
 
  • #3
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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...
 
Last edited:
  • #4
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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!!
 
  • #5
34
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what is the value of a^2 when a = -1?

Remember you square first then subtract!!
Careless mistakes can be lethal :frown:
 

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