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If sin x + cos x = -1/5, and 3/4*pi <= x <= pi, then find cos 2x.

 Dec 8, 2019
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If sin x + cos x = -1/5, and 3/4*pi <= x <= pi, then find cos 2x.

 

sin x  +  cos x  =  -1/5        square both sides

 

sin^2 x + 2sinxcosx  + cos^2x  = 1/25

 

1 + 2sin xcosx =  1/25

 

2sinxcosx =  -24/25

 

sin (2x)  =  -24/25

 

cos (2x)  = ±√ [ 1 - sin^2(2x)  ]

 

cos (2x)  = ±√[ 1 - (-24/25)^2 ] 

 

cos (2x)  = ±√[ 625 - 576]/ 25

 

cos (2x)  =  ± (7/25)

 

Since  x =    [ (3/4)pi , pi ]   then

 

arccos [ 7/25]  = x  ≈  1.287 rads

Then dividing by 2, x  ≈  .6435 rads......reject

 

arccos [ -(7/25) ]   = x  ≈  1.85 rads

Then dividing by 2, x    ≈  .927 rads   .....reject

 

And

 

arccos  [ 7/25]  =  2pi - 1.287  =  4.996 rads

Then dividing by 2, x ≈  2.498 rads.....accept

 

 

So cos 2x   =  7/25

 

cool cool cool

 Dec 8, 2019
edited by CPhill  Dec 8, 2019

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