If #costheta=-24/25# and #pi/2<theta<pi#, how do you find #sin2theta#?

1 Answer
Sep 28, 2016

#sin2theta = -336/25#

Explanation:

The restrictions mean that #theta# is in the second quadrant.

Hence, sine will be positive and cosine will be negative.

The expansion of #sin2theta# will be #2sinthetacostheta#, following the double angle identity. So, we need to determine the value of #sintheta#.

Recall that #costheta= "adjacent"/"hypotenuse"#, so #"adjacent" = -24# and #"hypotenuse" = 25#. By pythagorean theorem, we have that:

#o^2 + (-24)^2 = 25^2#, where the side opposite #theta = o#

#o^2 = 625 - 576#

#o^2 = 49#

#o = +-7#

However, we know that in quadrant 2, sine is positive, so the positive answer is the correct one.

Now, recall that #sintheta = "opposite"/"hypotenuse" = 7/25#

Therefore, we can calculate #sin2theta# as the following:

#sin2theta = 2sinthetacostheta = 2(7/25)(-24/25) = -336/25#

Hopefully this helps!