How do you use the half angle formula to determine the exact values of the sine, cosine, and tangent #-75^circ#?

1 Answer
Jul 26, 2018

#sin75^circ=(sqrt6+sqrt2)/4#
#cos75^circ=(sqrt6-sqrt2)/4#
#tan75^circ=2+sqrt3#

Explanation:

We know that ,

#color(red)((1)sin(x+y)=sinxcosy+cosxsiny#

#color(blue)((2)cos(x+y)=cosxcosy-sinxsiny#

#color(green)((3)tan(x+y)=(tanx+tany)/(1-tanxtany)#

We have ,

#color(brown)(75^circ=45^circ+30^circ#

#diamondsin75^circ=color(red)(sin(45^circ+30^circ) toApply(1)#

#=>sin75^circ=color(red)(sin45^circcos30^circ+cos45^circsin30^circ#

#=>sin75^circ=1/sqrt2*sqrt3/2+1/sqrt2*1/2#
#=>sin75^circ=(sqrt3+1)/(2sqrt2)#
#=>sin75^circ=(sqrt3+1)/(2sqrt2) xxsqrt2/sqrt2#
#=>sin75^circ=(sqrt6+sqrt2)/4#

#diamondcos75^circ=color(blue)(cos(45^circ+30^circ)toApply(2)#

#=>cos75^circ=color(blue)(cos45^circcos30^circ-sin45^circsin30^circ#

#=>cos75^circ=1/sqrt2*sqrt3/2-1/sqrt2*1/2#
#=>cos75^circ=(sqrt3-1)/(2sqrt2)#
#=>cos75^circ=(sqrt3-1)/(2sqrt2) xxsqrt2/sqrt2#
#=>cos75^circ=(sqrt6-sqrt2)/4#

#diamondtan75^circ=color(green)(tan(45^circ+30^circ)toApply(3)#

#=>tan75^circ=color(green)((tan45^circ+tan30^circ)/(1-tan45^circtan30^circ)#
#=>tan75^circ=(1+1/sqrt3)/(1-1/sqrt3)#
#=>tan75^circ=(sqrt3+1)/(sqrt3-1)#
#=>tan75^circ=(sqrt3+1)/(sqrt3-1)xx(sqrt3+1)/(sqrt3+1)#
#=>tan75^circ=(sqrt3+1)^2/((sqrt3)^2-1^2)#
#=>tan75^circ=(3+2sqrt3+1)/(3-1)#
#=>tan75^circ=(4+2sqrt3)/2#
#=>tan75^circ=2+sqrt3#

Note:
If it is #(-75^circ)# ,then take #-75^circ=-45^circ-30^circ .#
So,
#sin(-75^circ)=-sin75^circ=-((sqrt6+sqrt2)/4)#
#cos(-75^circ)=cos75^circ=(sqrt6+sqrt2)/4#

#tan(-75)^circ=-tan75^circ=-(2+sqrt3)#