Question 937985: hey there,
I'm doing some basic trig functions and would like to get some insight into this question i am struggling with.
let 0 <= delta < pi/2
(a) working from the unit circle, demonstrate that tan(pi - delta) = -tan delta
(b) hence, find the exact value of tan(pi - pi/6)
thanks,
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! let 0 <= delta < pi/2
(a) working from the unit circle, demonstrate that tan(pi - delta) = -tan delta
Draw a unit circle centered at (0,0)
Pick a point in QII ; Label the point (-x,y)
x is negative and y is positive
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That point is at pi-delta, eg. 180-30 = 150 degrees
tan(pi-delta) = y/-x = -y/x
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Draw a diameter from that point, thru (0,0) till it meets the circle in QIV
That point has coordinates (x,-y)
That point is delta below the x-ax or at -delta
The tangent is -y/x which is the negative of tan(pi-delta)
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(b) hence, find the exact value of tan(pi - pi/6) = -tan(pi/6) = -1/sqrt(3)
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Cheers,
Stan H.
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