given cosx=1/4, cscx<0 find cos2x then find tanx/2

Answer 1

cosx=1/4

sinx =+/-sqr(16-1)/4= +/-sqr15/4

but cscx<0 means the 3rd or 4th quadrant

cosx>0 means the 4th quadrant

so sinx <0

sinx =-sqr15/4

cos2x = cos^2(x) - sin^2(x) = 1/16 - 15/16 = - 14/16 = -7/8=-.875

or

cos2x = 2cos^2(x) -1 = 2((1/4)^2) -1 = 2/16 -1 = 1/8- 1 = - 7/8

sin2x = 2sinxcosx = 2(-sqr15)/4)(1/4) = -sqr15/8

tan2x = sin2x/cos2x = -sqr15/8)/(-7/8) = sqr15/7

check the answers

find x

= cos^-1(1/4) = about 75 1/2 degrees

cos2x = cos151 degrees =-.875 =- 7/8

tan2x = tan151 =about -.554 = about -sqr15/7 = about .553 close enough

Answer 2

Since csc x<0 and cos x>0, x is in quadrant IV.

arccos(.25) in quadrant 4 is 360-75.522488

You need to mess with this angle to get the answers.