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

Answer:

#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

Answer:

#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

Answer:

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#