Find sin2x and cos(x/2) given that cotx= -(3/4) and cosx < 0

Question

 Help please?

Answer 1

Recall cotx = cosx / sinx. Set y = cosx and z = sinx. Then we are given y / z = -(3 / 4). We also know (trig identity) that y^2+z^2=1. Two unknowns (y and z) and two equations means we can solve for y and z!

 

Solve y^2+z^2 = 1 for y^2: 

y^2 = 1 - z^2.

 

Square both sides of y / z = -(3 / 4) and substitute our equation for y^2:

y^2 / z^2 = 9 / 16

(1 - z^2 ) / z^2 = 9 / 16

1 / z^2 - 1 = 9 / 16

1 / z^2 = 25 / 16

z^2 = 16 / 25

z = plus or minus 4 / 5

 

Recall z = sinx. Since cosx<0, and cotx = -(3/4)<0, we see that sinx>0. Thus, z = sinx = 4/5.

 

Now we can solve for y via either equation:

 

y^2 + z^2 = 1

y^2 + (4/5)^2 = 1

y^2 + 16/25 = 1

y^2 = 9/25

y = plus or minus 3/5

 

Since cosx<0, we see that y = cosx = -3/5.

 

(Note that these values for cosx and sinx look good because they reside in the interval [-1,1]. Simple checks like this can save you from making silly mistakes!)

 

Now that we have found values for sinx and cosx, we use double and half-angle formulas to solve for sin2x and cos(x/2):

 

sin2x = 2sinxcosx = 2(4/5)(-3/5) = -24/25.

cos(x/2) = sqrt{ (1 + cosx)/2 } = sqrt{ (1-3/5)/2 } = sqrt{ 2/5/2 } = sqrt{ 1/5 }.