Asked by Joe
Given that SIN Theta = -1/sqrt(10) and theta is in quadrant 4, find sin2theta and cos2theta.
I'm not sure how to properly square the -1/sqrt10 and keep an exact answer other than decimal. Otherwise, my final answer is Sin2Theta = -.7255970691 and cos2theta = 1.632455533
I'm not sure how to properly square the -1/sqrt10 and keep an exact answer other than decimal. Otherwise, my final answer is Sin2Theta = -.7255970691 and cos2theta = 1.632455533
Answers
Answered by
Reiny
sketch a triangle in quadrant IV with opposite side 1 unit long, and the hypotenuse √10 units long.
The the adjacent side is 3 units
sinØ = -1/√10
cosØ = 3/√10 , remember the cosine is + in IV
sin 2Ø = 2sinØcosØ = 2(-1/√10)(3/√10) = -6/10 = -3/5
cos 2Ø = cos^2 Ø - sin^2Ø = 9/10 - 1/10 = 4/5
(I knew your answer was not correct since the cosine or sine of any angle cannot be bigger than 1. )
The the adjacent side is 3 units
sinØ = -1/√10
cosØ = 3/√10 , remember the cosine is + in IV
sin 2Ø = 2sinØcosØ = 2(-1/√10)(3/√10) = -6/10 = -3/5
cos 2Ø = cos^2 Ø - sin^2Ø = 9/10 - 1/10 = 4/5
(I knew your answer was not correct since the cosine or sine of any angle cannot be bigger than 1. )
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