SOLUTION: If cos(theta) = -3/7 with theta in quadrant II find the exact value of tan(theta - pi/3). I can't figure this out. Thank you.

Question 806041: If cos(theta) = -3/7 with theta in quadrant II find the exact value of tan(theta - pi/3). I can't figure this out. Thank you.
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
If cos(theta) = -3/7 with theta in quadrant II find the exact value of tan(theta - pi/3).
***
use x for theta
sin(x)=√(1-cos^2(x))=√(1-9/49)=√(40/49)=√40/7
tan(x)=sin(x)/cos(x)≈-√40/3
Identity: tan(a-b)=(tana-tanb)/(1+tana*tanb)
tan(x-π/3)=(tan(x)-tan(π/3))/(1+tan(x)*tan(π/3)
tan(π/3)=√3
tan(x-π/3)=(-√40/3-√3)/(1+(-√40/3*√3)(exact value)
..
check: (w/calculator)
cos(x)=-3/7 (in quadeant II)
reference angle≈64.62˚
terminal angle(x)≈ 115.38˚
x-π/3=115.38-60≈55.38˚
tan(55.38˚)≈1.449
Exact value as calculated:
=(-√40/3-√3)/(1+(-√40/3*√3)
≈-3.84/-2.65≈1.449

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