Question 856042: Given tan θ=-1/√3 find θ for 0≤θ<2π
If you could show how to answer this question that would be appreciated
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Given tan θ=-1/√3 find θ for 0≤θ<2π
If you could show how to answer this question that would be appreciated
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Draw a circle showing the 4 quadrants.
Since given tan θ is<0, reference angle θ must be in quadrant II or quadrant IV.
tan θ=sin θ/cos θ
if in quadrant II, cos<0, in quadrant IV, sin θ<0
in quadrant II, draw a reference right triangle with the opposite(y) leg=1, and the adjacent leg(x) leg=-√3
This will give you a hypotenuse=2
sin θ=opposite side/hypotenuse=1/2
so, θ=π/6 or 30˚
..
you can repeat this procedure in quadrant IV with the signs for opposite and adjacent legs of the reference right triangle reversed.
in this case, θ=11π/6 or 330˚
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