Find the exact value?

Question

cos*theta = 12/13

Find the exact value of:

a) sin (theta/2)

b) cos (theta/2)

Answer 1

Given
cosθ = 12/13
________

Find the exact value of:

a) sin(θ/2)

There is one solution whether θ is in the first quadrant or the fourth quadrant. This is because θ/2 will be in the first quadrant or the second quadrant. Either way sin(θ/2) is positive.

sin(θ/2) = √[(1 - cosθ)/2] = √[(1 - 12/13)/2] = √(1/26)
sin(θ/2) = 1/√26

b) cos(θ/2)

There will be two solutions depending on whether θ is in the first quadrant or the fourth quadrant. This is because θ/2 will be in the first quadrant where cos(θ/2) is positive or in the second quadrant where cos(θ/2) is negative.

cos(θ/2) = ±√[(1 + cosθ)/2] = ±√[(1 + 12/13)/2] = ±√(25/26)
cos(θ/2) = ±5/√26

Answer 2

cos(theta) = 12/13 >>> theta = 22.61985

sin(22.61985/2) = 0.19612

cos(22.61985/2) = 0.98058

Answer 3

cos x = 1 - 2*sin^2 x so sin^2 (x/2) = (1 - cos x)/2 = 1/26. Hence sin (x/2) = 1/ sqr 26 and cos (x/2) = 5/sqr 26 by Pythagoras.

Answer 4

a)
Use the half angle formulae for the sine:
sin(θ/2) = ± sqrt( ( 1 - cosθ) / 2)
= ± sqrt( ( 1 - 12/13) / 2)
= ± sqrt(1 / 26)
= ± 1 / sqrt(26)

There are two solutions because there exist two angles θ for which cosθ = 12/13:
θ₁ = 22.62°
θ₂ = 360° - θ₁ = 337.38°

b)
Use the half angle formulae for the cosine:
cos(θ/2) = ± sqrt( ( 1 + cosθ) / 2)
= ± sqrt( ( 1 + 12/13) / 2)
= ± sqrt(25 / 26)
= ± 5 / sqrt(26)

Answer 5

cos(theta)=12/13
therefore sin(theta)=sqrt(1-(12/13)^2)
=sqrt(1-(144/169))
=sqrt((169-144)/169)
=sqrt(25/169)
therefore sin(theta)=5/13
a)
we know that cos(theta)=1-2(sin^2(theta/2))
=> (12/13)=1-2(sin^2(theta/2))
=> 2sin^2(theta/2)=1-12/13=11/13
=> sin^2(theta/2)=11/26
therefore sin(theta/2)=sqrt(11/26)=0.65
b)
we also know that cos(theta)=2cos^2(theta/2)-1
=> 2cos^2(theta/2)=1+cos(theta)
=> 2cos^2(theta/2)=1+12/13=14/13
=> cos^2(theta/2)=7/13=0.734
=> cos(theta/2)=sqrt(7/13)
therefore cos(theta/2)=

Answer 6

Do you know your half-angle formulas? If not, go to http://mathworld.wolfram.com/Half-AngleFormulas.html