If cot A = 4/3, and sec<0, find cos A/2?

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answers:
Krishnamurthy say: cot (A) = 4/3, tan (A) = 3/4
sin (A) = 3/5, cos (A) = 4/5
sec (A) < 0 means you're in quadrant 3
because cos(A) and sin(A) have to be negative
if cot (A) is positive with sec(A) < 0
cos(A) = -4/5
cos(A/2) = √(1 + cos (A)) / 2)
= √((1 - 4/5) / 2)
= √((1/5) / 2)
= √(1/10)
= 1/√10
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Como say: tan A = 3/4_______3rd quadrant
cos A = - 4/5
cos A = 2 cos² (A/2) -- 1
- 4/5 + 1 = 2 cos² (A/2)
1/10 = cos² (A/2)
- 1/√10 = cos A/2_________3rd quadrant
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Johnathan say: You have a 3/4/5 reference triangle (3^2 + 4^2 = 5^2); sec(A) < 0, so you know you're in quadrant 3 because cos(A) and sin(A) have to be negative if cot(A) is positive with sec(A) < 0. You actually DON'T need sin(A), though. cos(A) = -4/5...

cos(A/2) = sqrt((1 + cos(A)) / 2)

= sqrt((1 - 4/5) / 2)

= sqrt((1/5) / 2)

= sqrt(1/10), or sqrt(10)/10. Final.
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khalil say: cotA > 0 → A in quad 1 or 3
sec A = 1/ cos A < 0 → A in quad 2 or 3
so
A in quad 3 → sin A < 0 and cos A < 0
1 + cot^2 A = 1/ sin^2 A
1 + (4/3)^2 = 1/ sin^2 A
sin A = ±3/5
sinA = - 3/5 ... accepted answer
cos^2 A = 1 - sin^2 A
cos^2 A = 1 - (-3/5)^2
cos A = ± 4/5
cos A = - 4/5 .....accepted answer
cos (A/2) = √ [( 1+ cos A) / 2]
cos (A/2) = √(1/10)
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Robert say: Since cot A = 4/3, tan A = 3/4.

sec A < 0, so cos A < 0, too.

Thus A is in the 3rd quadrant.

So cos A = -4/5.

Now use the half-angle formula and you re done.
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