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

##### 3 Answers

#### 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#

#### Explanation:

Identity:

If:

Then:

This is in the **II** quadrant.

Where the cosine is negative:

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

#### Explanation:

First let's look at the double angle formula:

This cannot be used in its current form since

Substitution of

Now, looking at the question, if we let

Finally, substituting in the value of

Hence, the value of

The question specifies an answer for the third quadrant and so