SOLUTION: Hi can someone help me with this problem please? THANK YOU! Find the exact value of cos(2 theta) cos(theta)=-15/17 and theta terminates in QII. a.-2

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Question 1077273: Hi can someone help me with this problem please? THANK YOU!
Find the exact value of cos(2 theta)
cos(theta)=-15/17 and theta terminates in QII.
a.-240/289
b.161/289
c.120/289
d.16/17

Found 2 solutions by htmentor, MathTherapy:
Answer by htmentor(1343)   (Show Source): You can put this solution on YOUR website!
Given that cos(t) = -15/17, we know the adjacent side is 15 (QII) and the hypotenuse is 17.
Let the opposite side by x. Then sin(t) = x/17
From the Pythagorean theorem we know that x^2 = 17^2 - 15^2
We can make use of the identity, cos(2t) = cos^2(t) - sin^2(t)
sin^2(t) = x^2/17^2 = (17^2-15^2)/17^2 = 1 - (15/17)^2
Therefore, cos(2t) = (-15/17)^2 - [1-(15/17)^2] = (450-289)/289 = 161/289
Ans: b

Answer by MathTherapy(10551)   (Show Source): You can put this solution on YOUR website!

Hi can someone help me with this problem please? THANK YOU!
Find the exact value of cos(2 theta)
cos(theta)=-15/17 and theta terminates in QII.
a.-240/289
b.161/289
c.120/289
d.16/17

------ Substituting



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