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