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

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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#