How do you find the value of sin(theta/2) given tantheta=2 and 0<theta<pi/2?

2 Answers
Jul 24, 2016

sin(theta/2)=sqrt((sqrt5-1)/(2sqrt5))

Explanation:

It is observed that in range 0<theta<pi/2 all trigonometric functions are positive. Further as theta is in first quadrant, so is theta/2 hence trigonometric ratios of teta/2 too are all positive.

Now as tantheta=2, sectheta=sqrt(1+tan^2theta)

= sqrt(1+2^2)=sqrt5

and hence costheta=1/sqrt5

Now, sin(theta/2)=sqrt((1-costheta)/2)

= sqrt((1-1/sqrt5)/2)=sqrt((sqrt5-1)/(2sqrt5))

Jul 24, 2016

:.sin(theta/2)={(sqrt5-1)sqrt(50+10sqrt5)}/20

Explanation:

Let us use the Identity : tantheta=(2tan(theta/2))/(1-tan^2(theta/2)

Letting, tan(theta/2)=t, tantheta=(2t)/(1-t^2).............(1).

Since, tantheta=2, we have, by (1),

2=(2t)/(1-t^2)rArr1-t^2=t, or, t^2+t-1+0

t=(-1+-sqrt(1^2-4*1*(-1)))/2=(-1+-sqrt5)/2

Let us note that, by data, 0<,theta<,pi/2rArr 0<,theta/2<,pi/4

:. theta/2 lies in the First Quadrant, in which, tan(theta/2)>0,

and, also, sin(theta/2)>0

This means that tan(theta/2)!=(-1-sqrt5)/2

:. tan(theta/2)=(sqrt5-1)/2rArr cot(theta/2)=2/(sqrt5-1)

Now, csc^2(theta/2)=1+cot^2(theta/2)=1+4/(sqrt5-1)^2

:.csc^2(theta/2)={(5-2sqrt5+1)+4}/(sqrt5-1)^2=(10-2sqrt5)/(sqrt5-1)^2

:.sin(theta/2)=+(sqrt5-1)/sqrt(10-2sqrt5)={(sqrt5-1)sqrt(10+2sqrt5)}/{sqrt(10-2sqrt5)sqrt(10+2sqrt5)}

={(sqrt5-1)sqrt(10+2sqrt5)}/sqrt(100-20)={(sqrt5-1)sqrt(10+2sqrt5)}/(4sqrt5)

={(sqrt5-1)(sqrt5)sqrt(10+2sqrt5)}/(4*5)={(sqrt5-1)sqrt(50+10sqrt5)}/20