suppose that sin0=-5/13 and that 0 is a quadrant III angle. what is the exact value of cos0
The "reference angle" that theta makes with the -x axis is sin^-1(5/13) . The cosine of that angle is 12/13.
(Think of a 5:12:13 right triangle)
Cosine is negative in the third quadrant.
The answer is -12/13
make a sketch of the right-angled triangle in quad III, with hypotenuse 13 and opposite 5.
By Pythagoras the adjacent would be 12 (the 5-12-13 triangle)
so cosØ = -12/13 ( by CAST, in III the cosine is negative)
To find the exact value of cos(0), we can use the Pythagorean identity, which states that sin^2(θ) + cos^2(θ) = 1. Since we know the value of sin(0) is -5/13, we can substitute it into the equation:
(-5/13)^2 + cos^2(0) = 1
25/169 + cos^2(0) = 1
cos^2(0) = 1 - 25/169 = 144/169
Taking the square root of both sides, we get:
cos(0) = ± √(144/169)
Given that 0 is a quadrant III angle, the cosine value will be negative. So, we can write:
cos(0) = -√(144/169) = -12/13
Therefore, the exact value of cos(0) in this scenario is -12/13.