Question 802602: Good day,
My question is that I could use some assistance with is (thanks in advance!):
Given that theta = 17pi/6 use the period of the functions as well as the special angles to find the exact values (does this mean radians?) of the sin, cos and tan of theta.
Found 2 solutions by lwsshak3, Math1337:
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Given that theta = 17pi/6, find the exact values of the sin, cos and tan of theta.
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What you need to know is the reference angle of given angle theta=17π/6, and which quadrant it is in.
17π/6 is like rotating counter-clockwise around the unit circle 16π plus an additional π/6, which places the reference angle of π/6 in quadrant I.
sin(17π/6)=sin(π/6)=1/2
cos(17π/6)=cos(π/6)=√3/2
tan(17π/6)=tan(π/6)=sin/cos=1/√3=√3/3
Answer by Math1337(1) (Show Source):
You can put this solution on YOUR website! I challenge the previous answer's correctness, which states as follows:
Given that theta = 17pi/6, find the exact values of the sin, cos and tan of theta.
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What you need to know is the reference angle of given angle theta=17π/6, and which quadrant it is in.
17π/6 is like rotating counter-clockwise around the unit circle 16π plus an additional π/6, which places the reference angle of π/6 in quadrant I.
sin(17π/6)=sin(π/6)=1/2
cos(17π/6)=cos(π/6)=√3/2
tan(17π/6)=tan(π/6)=sin/cos=1/√3=√3/3
The above solution is incorrect.
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Correct Answer:
If theta = 17π/6, then when finding the quadrant,
The mistake above is in finding the correct quadrant of the unit circle.
Rotating 17π/6 is NOT "like rotating counter-clockwise around the unit circle 16π plus an additional π/6."
It is rotating 2π then an additional 5π/6, or rotating 3π minus π/6, which would place it in quadrant II.
Thus,
QED
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