If cos(t)=-7/8 with pi<=t<=3pi/2, how do you find the value of cos(t/2)?

3 Answers
Apr 8, 2018

cos(t/2)=-1/4

Explanation:

"using the "color(blue)"half angle formula"

•color(white)(x)cos(t/2)=+-sqrt((1+cost)/2)

pi<=t<=(3pi)/2rArrpi/2<=t/2<=(3pi)/4

rArrcos(t/2)=-sqrt((1-7/8)/2

color(white)(rArrcos(t/2))=-sqrt(1/16)=-1/4

Apr 8, 2018

cos(t/2)=-1/4

Explanation:

Identity:

color(red)bb(cos(x/2)=sqrt(1/2(1+cosx)

cos(t/2)=sqrt(1/2(1+(-7/8))

cos(t/2)=sqrt(1/2(1/8))

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \=sqrt(1/16)=+-1/4

If:

pi <= t <= (3pi)/2

Then:

t/2

pi/2 <=t/2 <= (3pi)/4

This is in the II quadrant.

Where the cosine is negative:

cos(t/2)=-1/4

Apr 8, 2018

The use of the some trigonometric identities is the key to solving this question.

Explanation:

First let's look at the double angle formula:
cos(2x) = cos^2(x) - sin^2(x)

This cannot be used in its current form since sin^2(x) is present. But, rearranging the pythagorean identity allows for the sin^2(x) to be replaced in terms of cos(x) :
sin^2(x) + cos^2(x) = 1
sin^2(x) = 1 - cos^2(x)

Substitution of sin^2(x) into double angle formula:
cos(2x) = cos^2(x) - (1 - cos^2(x))
cos(2x) = 2cos^2(x) - 1

Now, looking at the question, if we let t = 2x and t/2 = x it changes the double angle formula to a form which can be directly used:
cos(t) = 2cos^2(t/2) - 1

Finally, substituting in the value of cos(t) and rearranging the equation yields an answer:
-7/8 = 2cos^2(t/2) - 1
1/8 = 2cos^2(t/2)
1/16 = cos^2(t/2)
1/4 = cos(t/2)

Hence, the value of cos(t/2) is +-1/4 .
The question specifies an answer for the third quadrant and so
cos(t/2) = -1/4