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Angle θ is in quadrant II and sin θ \({5 \over 13}\) . Determine an exact value for cos 2θ.

 

Since angle θ is in quadrant II for cos 2θ, I used the CAST rule and I inquired that this value must be negative.

 

I also used the double angle identity of cos 2A = cos\({^2}\)A - sin\({^2}\)A.

 

If sin θ = \({5 \over 13}\), I subbed it in for A and multiplied it by 2.

 

I got - cos (\({10 \over 26}\)), but the answer in the textbook is \({119 \over 169}\) . How should I solve this?

 Feb 22, 2019
 #1
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We have a 5 - 12 - 13   right triangle in Q2

The cosine is negative  here  so it =   -12/13

 

cos 2 θ  =   cos^2  θ   -  sin^2  θ

 

So

 

cos ^2  θ  =  (-12/13)^2  =    144/169

 

And

 

sin^2  θ  =  ( 5/13)^2 =   25 /169

 

So

 

cos 2 θ  =   144/169  -  25/169    =      119 / 169     !!!!

 

 

 

cool cool cool

 Feb 22, 2019

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