find sin theta/2 and cos theta/2 for sintheta=2/7

Answer 1

Hello Jacob

find sin theta/2 and cos theta/2 for sintheta=2/7

Given Sin θ = 2/7

Sin θ = Sin(θ/2+θ/2) =2 Sinθ/2 Cosθ/2 =2/7

so

Sinθ/2 Cosθ/2= 1/7

But Sinθ = √ (1- Cos² θ )

hence Sinθ/2 = √ (1- Cos² θ/2 )

√ (1- Cos² θ/2 ) ( cos θ/2) = 1/7

Squaring both sides

(1- Cos² θ/2 ) Cos²θ/2 =1/49

Solve this for Cosθ/2

Then find Sinθ/2

1. Another method using geometry

Answer 2

Let angle theta = T

sin T = 2/7

opposite is 2 and hypotenuse is 7

half angle formula say: sin(T/2) = +or- sqrt[( 1 - cos T)/2]

cos(T/2) = +or- sqrt[( 1 + cos T)/2]

depending where the angle is...

This can be in quadrant #1 or quadrant #2 where sine is positive.

If quadrant 1:

then pythagorean says adjacent sqrt( 49 - 4) = sqrt(45) = 3 * sqrt(5)

so cos T = 3*sqrt(5)/7

in quadrant 1, cosine is positive, so

sin (T/2) = sqrt[(1 - cos(T))/2]

= sqrt ( (1/2) - (1/2) cos(T) ]

= sqrt( (1/2) - (1/2) (1/7)(3 * sqrt(5)))

= sqrt( (1/2) - (3/14) sqrt(5)))

= sqrt( (7/14) - (3/14) sqrt(5))) <-- common denominator

= sqrt ( (7 - 3 * sqrt(5)) / 14 )

= sqrt ( 14 ( 7 - 3 * sqrt(5)))/ 14 <--- rationalizes the denominator

cos(T/2) is the same, except the sign changes: sqrt ( 14 ( 7 - 3 * sqrt(5)))/ 14

If quadrant 2, the cosine is negative so cos T = -3*sqrt(5)/7

So the sin(T/2) shall have the same answer but the cosine is -sqrt ( 14 ( 7 - 3 * sqrt(5)))/ 14