Given tan(a)=7/sqrt(32) and (a) is in QIII, find the exact value of tan(a/2).

Question

No decimals are allowed in the answer.

Answer 1

Drawing the reference triangle in quadrant III, we have horizontal leg -√32, vertical leg -7, hypotenuse 9, using the pythagorean formula. From this we can express sin A (I prefer capital letters for the angle) = -7/9 and cos A = -√32/9.

 

Now you are in a position to use a half-angle formula. Simplify your result but avoid decimals, as directed.

Answer 2

tan=opposite/adjacent

draw a triangle in quadrant lll

the legs are 7 and √32=4√2

find the hypotenuse

H^2=7^2+(4√2)^2

H^2=49+32=81

H=√81=9

the hypotenuse is 9

the adjacent is 4√2

cos=adjacent/hypotenuse

cos=-4√2/9

tan(θ/2)=±√[(1-cosθ)/(1+cosθ)]

tan(θ/2)=±√[(1+4√2/9)/(1-4√2/9)]

tan(θ/2)=±√([9+4√2]/9)/[9-4√2]/9)

tan(θ/2)=±√[(9+4√2)/(9-4√2)]

multiply numerator and denominator by 9+4√2

tan(θ/2)=±√(81+72√2+32)/(81-32)

tan(θ/2)=±√(113+72√2)/(49)

tan(θ/2)=±[√(113+72√2)]/7