Given cos a=-7/25 in quadrant II , and sin b=-12/13, in quadrant IV , find cos(a+b) and sin2a
sin a = 24/25
cos b = 5/13
now apply addition formulas
Thanks but how did you find that? My numbers came out completely different -_-
If a is in QII, sin is positive, cos is negative.
7^2 + 24^2 = 25^2, so it's a 7-24-25 triangle
If b is in QIV, sin is negative, cos is positive.
5^2 + 12^2 = 13^2, so it's a 5-12-13 triangle.
what did you do to find the missing sin/cos values?
To find cos(a+b), we can use the cosine angle addition formula which states that cos(a+b) = cos(a)cos(b) - sin(a)sin(b).
Since cos a = -7/25, we know that adjacent side is -7 and the hypotenuse is 25. Using the Pythagorean theorem, we can find the opposite side of the triangle: sin a = sqrt(hypotenuse^2 - adjacent^2) = sqrt(25^2 - (-7)^2) = sqrt(625 - 49) = sqrt(576) = 24.
Similarly, for sin b = -12/13, we know that the opposite side is -12 and the hypotenuse is 13. Using the Pythagorean theorem, we can find the adjacent side: cos b = sqrt(hypotenuse^2 - opposite^2) = sqrt(13^2 - (-12)^2) = sqrt(169 - 144) = sqrt(25) = 5.
Now, we have the values of sine and cosine for both angles. Plugging these values into the cosine angle addition formula, we get cos(a+b) = (cos a)(cos b) - (sin a)(sin b) = (-7/25)(5) - (24)(-12/13) = -35/25 + 288/25 = 253/25.
So, cos(a+b) = 253/25.
Moving on to sin2a, we can use the double-angle formula for sine, which states that sin(2a) = 2sin(a)cos(a).
Using the values we already determined, sin2a = 2(sin a)(cos a) = 2(24/25)(-7/25) = -336/625.
So, sin2a = -336/625.