My Math Forum Finding the y-coordinate given an equation

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 September 24th, 2017, 09:20 AM #1 Newbie   Joined: Sep 2017 From: Daly City Posts: 1 Thanks: 0 Finding the y-coordinate given an equation I've been stuck on this trying to solve it on my own and with the textbook instructions but the lesson isn't in depth enough to help me solve this. I am given the equation cosθ + sin^2θ = -5/16 I know sin^2θ + cos^2θ = 1, but my cosθ isn't squared. I've tried squaring everything but then I end up with sin^4θ and more confusion... I got the question wrong so many times I can't try anymore but I got a window with a message saying: In a right​ triangle, with legs a and b and hypotenuse​ c, the Pythagorean Theorem states that a squared plus b squared equals c squareda2+b2=c2. Use the given equation with the Pythagorean Theorem to write an equation using only one of the trigonometric functions. Remember that cos(pi) = -1 and sin(3pi/2) = -1 I've run out of different ways to attempt this. Where can I go from here?
 September 24th, 2017, 09:52 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,767 Thanks: 1422 $\cos{t} + (1-\cos^2{t}) = -\dfrac{5}{16}$ $0 = \cos^2{t} - \cos{t} - \dfrac{21}{16}$ $a = 1$, $b = -1$, $c = - \dfrac{21}{16}$ $\cos{t} = \dfrac{1 \pm \sqrt{1 + \frac{21}{4}}}{2} = \dfrac{2 \pm 5}{4}$ for $0 \le t <2\pi$ ... $\cos{t} = \dfrac{7}{4}$ has no solution $\cos{t} = -\dfrac{3}{4}$ $t = \arccos\left(-\dfrac{3}{4}\right)$ $t = 2\pi - \arccos\left(-\dfrac{3}{4}\right)$ Thanks from Country Boy

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