Question 959220: Find exact Value of tan Beta/2, Given that tanBeta = square root 5/2 and pi < Beta < 3pi/2
So I think Beta/2 is in quadrant 3 making it positive but I tried square root of 5 as the answer and thats wrong please help!!!!
Found 2 solutions by lwsshak3, ikleyn:
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Find exact Value of tan Beta/2, Given that tanBeta = square root 5/2 and pi < Beta < 3pi/2
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use x for Beta
tanx=√5/2 (In quadrant III)
hypotenuse of reference right triangle in quadrant III=√(√5)^2+2^2)=√(5+4)=3
sinx=√5/3
cosx=2/3
tan(x/2)=sinx/(1+cosx)
tan(x/2)=√5/3/(1+2/3)=√5/3/5/3=√5/5
Check:
tanx=√5/2
x=48.1897˚
x/2≈24.0948˚
tan(x/2)≈tan(24.0948)≈0.4472
exact value=√5/5≈0.4472
Answer by ikleyn(53875)  (Show Source):
You can put this solution on YOUR website! .
Find exact Value of tan(Beta/2), Given that tan(Beta) = square root 5/2 and pi < Beta < 3pi/2
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Calculations in the post by @lwsshak3 are fatally and totally incorrect.
It is enough to notice that Beta is in QIII (given), hence, Beta/2 is in QII,
so tan(Beta) must be negative, while @lwsshak3 gives a positive number as the answer.
His checking procedure also is wrong.
Below is my correct solution.
use x for Beta
tan(x) = √5/2 (given, so BETA is in QIII)
hypotenuse of reference right triangle in quadrant III = √(√5)^2+2^2)=√(5+4)=3
sin(x) = -√5/3 in QIII (negative)
cos(x) = -2/3 in QIII (negative)
tan(x/2) = sin(x)/(1+cos(x))
tan(x/2) = (-√5/3)/(1-2/3) = (-√5/3)/(1/3) = -√5/1 = -√5.
Check:
tan(x) = √5/2 in QIII
x = 48.19° + 180° = 228.19°
x/2 ≈ 114.095°
tan(x/2) ≈ tan(114.095°) ≈ -2.236
exact value = -√5 ≈ -2.236
Solved correctly and checked in a right way, too.
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