Okay also, let me show one way to do it. This way may not be the way shown in your book.Originally Posted bymathAna1ys!5

So the angles are greater than 90 degrees. The first quadrant is from 0 to 90 degrees only. Hence, find an angle from 0 to 90 degrees that will represent the given large degree, meaning, get the acute angle based from the x-axis. Like plot the given large degree on the x,y axes setup.

Then use the given trig function on the equivalent acute angle.

Your tan(315deg).

315 degrees is in the 4th quadrant. It is 45deg less than a complete 360deg.

So, using the given trig function, tan(315deg) = tan(45deg) in the 4th quadrant.

Tangent in the 4th quadrant is negative. So, tan(315deg) = -tan(45deg).

cos(562deg)

562 -360 = 202

202 -180 = 22, so 562deg is in the 3rd quadrant. It is 22deg in the 3rd quadrant.

Cosine is negative in the 3rd quadrant, hence, cos(562deg) = -cos(22deg).

sin(-792deg)

Negative angle means it is meaured clockwise.

792 -(360 +360) = 792 -720 = 72.

If you plot that on the x,y axes, you start from the positive side of the x-axis, then go clockwise 360deg---4th quadrant, 3rd quadrant, 2nd quadrant, 1st quadrant---, another 360deg, until you stop at the the 4th quadrant.

So, -792deg is 72deg in the 4th quadrant.

Sine is negative in the 4th quadrant, so, sin(-792deg) = -sin(72deg).

sec(-210deg)

Go clockwise, -210deg stops in the 2nd quadrant. It is 210-180 = 30deg from the negative side of the x-axis.

Secant, or 1/cos, is negative in the 2nd quadrant, so,

sec(-210deg) = -sec(30deg).

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Do you know why any trig function is positive or negative in any of the four quadrants?