Given Csc(x)=8; how do you find sin x/2, cos x/2, tan x/2?

Question

12573 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-02T10:08:32

Two solutions - (A) if $cos(x/2)=1/2(3sqrt2+sqrt14)$ , $sin(x/2)=1/2(3sqrt2-sqrt14)$ and $tan(x/2)=8-3sqrt7$

and (B) if $cos(x/2)=1/2(3sqrt2-sqrt14)$ , $sin(x/2)=1/2(3sqrt2+sqrt14)$ and $tan(x/2)=8+3sqrt7$

As $cscx=8$ , $sinx=1/cscx=1/8$ and as $sinx>0$ , we have $0 < x < pi$ and

$0 < x/2 < pi/2$ and hence $x/2$ lies on Q1 and all trigonometric ratios are positive.

As $sinx=1/8$ ,

$cosx=sqrt(1-(1/8)^2)=sqrt(1-1/64)=+-sqrt63/8$

and as $cos2A=2cos^2A-1=1-2sin^2A$ , we have

$cosA=sqrt((1+cos2A)/2)$ and $sinA=sqrt((1-cos2A)/2)$

Hence $cos(x/2)=sqrt((1+cosx)/2)$ i.e.

$sqrt((1+-sqrt63/8))/2=sqrt((8+-sqrt63)/16)=sqrt(8+-3sqrt7)/4=1/2(3sqrt2+-sqrt14)$

and $sin(x/2)=sqrt((1-cosx)/2)$ i.e.

$sqrt((1+-sqrt63/8))/2=sqrt((8+-sqrt63)/16)=sqrt(8+-3sqrt7)/4=1/2(3sqrt2∓sqrt14)$

Note that (A) if $cos(x/2)=1/2(3sqrt2+sqrt14)$ , $sin(x/2)=1/2(3sqrt2-sqrt14)$ and $tan(x/2)=sqrt(8-sqrt63)/sqrt(8+sqrt63)=8-3sqrt7$

and (B) if $cos(x/2)=1/2(3sqrt2-sqrt14)$ , $sin(x/2)=1/2(3sqrt2+sqrt14)$ and $tan(x/2)=sqrt(8+sqrt63)/sqrt(8-sqrt63)=8+3sqrt7$