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
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. )
To find sin(2θ) and cos(2θ), we can use the double angle identities:
sin(2θ) = 2sin(θ)cos(θ)
cos(2θ) = cos²(θ) - sin²(θ)
First, let's find sin θ and cos θ from the given information SIN θ = -1/√10 and θ is in quadrant 4.
Since θ is in quadrant 4, sine is negative in this quadrant.
sin θ = -1/√10 (given)
cos θ = √(1 - sin²θ)
= √(1 - (-1/√10)²)
= √(1 - 1/10)
= √(9/10)
= 3/√10
Now that we have sin θ and cos θ, we can find sin(2θ) and cos(2θ).
sin(2θ) = 2sin(θ)cos(θ)
= 2(-1/√10)(3/√10)
= -6/10
= -3/5
cos(2θ) = cos²(θ) - sin²(θ)
= (3/√10)² - (-1/√10)²
= 9/10 - 1/10
= 8/10
= 4/5
Therefore, sin(2θ) = -3/5 and cos(2θ) = 4/5.