cos(θ)=-((3)/(7)) where pi<theta <3pi/2 find value sin2 theta
1. Get the inverse of each side in cos(θ)=-((3)/(7)).
Cos^-1 (3/7) = theta.
2. From trig identities, we know sin2θ = 2sinθcosθ
Substitute that in for sin2θ and just solve since you figured out θ in part 1 and you were given what cosθ equals, which is 3/7.
Oh, I forgot to add the negative sign with 3/7.
To find the value of sin^2(theta), we first need to determine the value of sin(theta).
Given that cos(theta) = -3/7 and pi < theta < (3pi/2), we can use the Pythagorean Identity to find the value of sin(theta).
The Pythagorean Identity states: sin^2(theta) + cos^2(theta) = 1.
First, let's solve for cos^2(theta):
cos^2(theta) = (-3/7)^2 = 9/49
Now, let's substitute the value of cos^2(theta) in the Pythagorean Identity:
sin^2(theta) + 9/49 = 1
To solve for sin^2(theta), we rearrange the equation:
sin^2(theta) = 1 - 9/49
Now, we can find the value of sin^2(theta):
sin^2(theta) = (49/49) - (9/49) = 40/49
Therefore, sin^2(theta) = 40/49.