If cos ⁡x=−1/2, and 90°<x<180°, find Sin X and Cot X

Question

With work please I don’t get it T-T

Answer 1

Hi,

If x is between 90 and 180, that means you're in quadrant 2. We can use this information to build a right triangle in Q2. Draw a line at whatever angle you want from the origin to somewhere in the second quadrant. Label an angle from that line down to the x-axis as "X". Complete the triangle by drawing a vertical line from the end of the angled line down to the x-axis.

Since cos(X) = -1/2, the adjacent leg to the angle (that is, the length along the x-axis) will be -1. The hypotenuse of this triangle will have a length of 2. By the Pythagorean Theorem, the height of this triangle will be sqrt(3). We've now got a fully defined triangle in our second quadrant.

sin(X) will be the opposite (sqrt(3)) over the hypotenuse (2). So sin(X) = sqrt(3)/2. This makes sense because sin is positive in Q2.

cot(X) will be the adjacent (-1) over the opposite (sqrt(3)). So cot(X) = -1/sqrt(3) or -sqrt(3)/3. This also makes sense because cot is negative in Q2.

It's a bummer I can't draw the triangle, but I hope my description helps you picture what's going on.

Answer 2

The magnitude of the sin and cot are respectively sqrt(3)/2 and sqrt(3).

In quadrant 2 the sin is + and the cot is -.

You might like to remember the mnemonic "AllSinTanCos" which tells you which trig function is positive in quadrant 1,2,3,4.