How do you find the value?

cos alpha/2 if tan alpha= 40/9, (180 < alpha<270)

2 Answers
Feb 19, 2018

cos(alpha/2)=-4/sqrt41

Explanation:

Given:
(180ltalpha<270)
indicating that cosalpha is negative

tanalpha=40/9
opposite side is 40
adjacent side is 9
hypotenuse will be 41,
because
sqrt(9^2+40^2)=sqrt(81+1600)=sqrt1681=41
cosalpha="(adjacent side)/(hypotenuse)"
cosalpha=-9/41
cos(alpha/2)=+-sqrt((1+cosalpha)/2)=+-sqrt((1-9/41)/2
=+-sqrt((41-9)/(41xx2))=+-sqrt(32/2xx1/41)=+-sqrt(16)/sqrt(41)

cos(alpha/2)=+-4/sqrt41

As mentioned, 180^@ltalphalt270^@,
it follows that
180/2ltalpha/2<270/2

90ltalpha/2<135, where cos(alpha/2) is negative

Hence,
cos(alpha/2)=-4/sqrt41

Feb 19, 2018

Start with the identity:

tan^2(alpha) +1 = sec^2(alpha)

Substitute sec^2(alpha)= 1/cos^2(alpha)

tan^2(alpha)+ 1= 1/cos^2(alpha)

Multiply both sides by cos^2(alpha)/(tan^2(alpha) + 1):

cos^2(alpha) = 1/(tan^2(alpha) + 1)

Use the square root operation on both sides:

cos(alpha) = +-sqrt(1/(tan^2(alpha) + 1))

We are told that 180^@ < alpha < 270^@, therefore we choose the negative value:

cos(alpha) = -sqrt(1/(tan^2(alpha) + 1))

Add 1 to both sides:

1+ cos(alpha) = 1-sqrt(1/(tan^2(alpha) + 1))

Multiply both sides by 1/2:

(1+ cos(alpha))/2 = (1-sqrt(1/(tan^2(alpha) + 1)))/2

Use the square root operation on both sides:

+-sqrt((1+ cos(alpha))/2) = +-sqrt((1-sqrt(1/(tan^2(alpha) + 1)))/2)

Substitute +-sqrt((1+ cos(alpha))/2)= cos(alpha/2)

cos(alpha/2) = +-sqrt((1-sqrt(1/(tan^2(alpha) + 1)))/2)

From 180^@ < alpha < 270^@ we derive 90^@ < alpha/2 < 135^@ and conclude that the cosine function is negative within the specified domain:

cos(alpha/2) = -sqrt((1-sqrt(1/(tan^2(alpha) + 1)))/2)

Substitute tan^2(alpha) = (40/9)^2:

cos(alpha/2) = -sqrt((1-sqrt(1/((40/9)^2 + 1)))/2)

I used WolframAlpha to simplify the above into an exact form:

cos(alpha/2) = -(4sqrt41)/41