Given that sin = 3/5 and lies in quadrant II, find the following value.
cos
To find the value of cos, we can use the identity:
sin²θ + cos²θ = 1
We are given that sinθ = 3/5. Since θ lies in quadrant II, which is to the left of the y-axis, we know that cosθ will be negative.
Using the identity sin²θ + cos²θ = 1, we can rearrange the equation to solve for cos²θ:
cos²θ = 1 - sin²θ
cos²θ = 1 - (3/5)²
= 1 - (9/25)
= 16/25
Taking the square root of both sides to solve for cosθ:
cosθ = ± √(16/25)
cosθ = ± (4/5)
Since θ lies in quadrant II where cosθ is negative, the value of cosθ is:
cosθ = -4/5
To find the value of cos, we can use the Pythagorean identity:
sin^2θ + cos^2θ = 1
Since sin = 3/5, we can substitute this value in:
(3/5)^2 + cos^2θ = 1
9/25 + cos^2θ = 1
Now, let's solve for cos^2θ:
cos^2θ = 1 - 9/25
cos^2θ = 25/25 - 9/25
cos^2θ = 16/25
Taking the square root of both sides to find the value of cos:
cosθ = ± √(16/25)
Since θ lies in quadrant II, where the cosine is negative, we choose the negative square root:
cosθ = - √(16/25)
cosθ = -4/5
Therefore, the value of cos is -4/5.
Isn't cosine negative in quadrant II?
memorize this:
Sin^2 X + Cos^2 X=1
find cosine X, you know sin X as 3/5