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