What is cos 2theta for sin theta = -2/13 in 180 degs < theta < 270 degs?
sinØ = -2/13 and we are told Ø is in III
so y = -2, r = 13
x^2 + y^2 = r^2
x^2 + 4 = 169
x= ±√165 , but in III x = -√165
cosØ = -√165/13
cos 2Ø = cos^2 Ø - sin^2 Ø
= 165/169 - 4/169
= 161/169
To find the value of cos(2θ) given sin(θ) = -2/13 and 180° < θ < 270°, we can follow these steps:
1. Draw a right triangle in the third quadrant (where sin is negative) with a reference angle θ and a vertical side of -2 and a hypotenuse of 13. Keep in mind that the horizontal side will be negative as well.
2. Now, we can find the cosine of θ. Since sin(θ) = -2/13, we can use the Pythagorean theorem to find the length of the horizontal side. The formula is:
sin²(θ) + cos²(θ) = 1
Plugging in the given values:
(-2/13)² + cos²(θ) = 1
4/169 + cos²(θ) = 1
cos²(θ) = 1 - 4/169
cos²(θ) = 169/169 - 4/169
cos²(θ) = 165/169
Taking the square root of both sides:
cos(θ) = ±√(165/169)
However, since we are working in the third quadrant (where cosine values are negative), we take the negative sign:
cos(θ) = -√(165/169)
3. Lastly, we need to find cos(2θ). Using the double-angle identity for cosine:
cos(2θ) = cos²(θ) - sin²(θ)
Plugging in the values we found earlier:
cos(2θ) = (-√(165/169))² - (-2/13)²
cos(2θ) = (165/169) - (4/169)
cos(2θ) = 161/169
Therefore, cos(2θ) equals 161/169.