Given sin(theta) = -3/5, and 180° < theta < 270°, determine the exact value of cos (theta)/2.

Question

Given sinθ = -3/5, and 180° < θ < 270°, determine the exact value of cosθ/2

Answer 1

Hi Trixie. The sine value -3/5 falls in quadrant III, because sine is negative in III, and quadrant III has degree value between 180 and 270 degrees.

Because the question asks for the cosine value, we get -4/5. This comes from third side of the right triangle in quadrant III, which is 4, which is the adjacent side, and the hypotenuse, which is 5.

Using the half angle formula for cosine, we get +/- the square root of 1 + -4/5/2, or 1-4/5/2, or +/- the square root of 1/5/2, = +/- the square root of 1/10.

Therefore, the answer is both the positive and negative square root of 1/10.

Answer 2

Hi Trixie,

First, we know the angle theta is in Quadrant III because theta is between 180 and 270 degrees, and this also works with the sin of theta being negative.

We need to find the cosine of theta, which will be negative for Quadrant III angles.

Think about SOHCAHTOA, and what sin and cosine are referring to.

sin(theta) = opp/hyp

cos(theta) = adj/hyp

(in other words)

sin(theta) = vertical component / hypotenuse

cos(theta) = horizontal component / hypotenuse

For this problem, we can just use a variable H for hypotenuse, and it will cancel out later. It doesn't matter what H is because this problem is not asking about the radius or distance of anything, it is only asking about angles.

I'll call the vertical component "y" and the horizontal component "x".

sin(theta) = y/H

sin(theta) is given as -3/5

-3/5 = y/H

Simply use the numbers in the fraction for y and H:

y = -3

H = 5

Next, we need to understand that we're looking at a triangle in Quadrant III.

Think about this angle in Quadrant III and how it forms a triangle with its horizontal and vertical components and its radius or hypotenuse.

The triangle has height of -3, and hypotenuse of 5.

Let's solve for the horizontal component of this triangle, because that can be used to determine cosine of theta!

In a triangle with one side length 3 and the hypotenuse 5, recall the special 3-4-5 triangle, and quickly know that the horizontal component is 4. Or if you don't remember the special triangle, you can use Pythagorean theorem to solve. 3² + x² = 5² (9 + 16 = 25, so x=4)

We know from the unit circle and Quadrant III that the horizontal component is negative.

Therefore, x = -4

Still we have:

y = -3

H = 5

Now, cos(theta) = x/H

cos(theta) = -4/5

Lastly, the problem asks for cos(theta)/2, so divide -4/5 by 2.

The answer is -2/5.

Hope this helps! :)