# Solving $\cos^2{\theta}-\sin{\theta} = 1$

$$\cos^{2}{\theta}-\sin{\theta} = 1, \quad\theta\in[0^{\circ}, 360^{\circ}]$$

• What have you done so far? We're more than willing to help get you through what has you stumped, but we like to see some work of your own. – Michael Dyrud Aug 23 '15 at 2:44
• Hint: $\sin^2\theta + \cos^2\theta = 1$. – vadim123 Aug 23 '15 at 2:45
• Is that $360^\circ$ or $360$ radians? – Rocket Man Aug 23 '15 at 2:50
• It is 360 degree AJ Stas – barryedx Aug 23 '15 at 2:52
• Also I am working on paper and I am not sure how to post math related questions with all the signs; eg theta – barryedx Aug 23 '15 at 2:54

$\cos^{2}(\theta) - \sin(\theta) = \cos^{2}(\theta) + \sin^{2}(\theta)$, (using Pythagorean Identity)

$\sin^{2}(\theta) + \sin(\theta) = 0$,

let $x = \sin(\theta)$

$x^{2}+ x = 0$

$x=\frac{-1 \pm \sqrt{1^{2}- 4(1)(0)}}{2}$, (Quadratic Formula)

$x = 0$, $x = -1$

substitute $\sin(\theta)$ back in and solve

$0 = \sin(\theta) \Rightarrow \theta = 0^{\circ},180^{\circ},360^{\circ}$

$-1 = \sin(\theta) \Rightarrow \theta = -270^{\circ}$

$\theta = \{0^{\circ},180^{\circ},270^{\circ},360^{\circ}\}$.

• You really don't need the quadratic formula to solve the equation $x^2 + x = 0$. – anomaly Aug 23 '15 at 3:38
• @anomaly I am glad Tim used it, because it's more general. OP may not have been aware (as OP couldn't solve this alone) – Alec Teal Sep 11 '15 at 3:56
• Come on! If someone doesn't recognize that the roots of $x^2+x=0$ are 0 and -1 becsuse $x^2+x = x(x+1)$, then, in my opinion, their understanding is weak. This was a blind application of the quadratic formula where it was absolutely not needed. – marty cohen Sep 11 '15 at 5:03

Notice, we have

$$\cos^2\theta-\sin\theta=1$$ $$1-\sin^2\theta-\sin\theta=1$$ $$-\sin^2\theta-\sin\theta=0$$ $$\sin^2\theta+\sin\theta=0$$ $$\sin\theta(\sin\theta+1)=0$$ $$\sin\theta=0\iff \theta=n(180^\circ)$$ Where, $n$ is any integer.

But for the given interval $[0^\circ, 360^\circ]$, substituting $n=0, 1, 2$, we get

$$\theta=0^\circ, 180^\circ, 360^\circ$$ Now, $$\sin\theta+1=0$$ $$\sin\theta=-1\iff \theta=2n(180^\circ)-90^\circ$$ But for given interval $[0^\circ, 360^\circ]$, substituting $n=1$ we get

$$\theta= 270^\circ$$ Hence, we have $$\color{red}{\theta}=\left\{\color{blue}{0^\circ, 180^\circ, 270^\circ, 360^\circ} \right\}$$

• Please explain how the negative sign turn to positive in step no. 3, thanks – barryedx Aug 23 '15 at 2:59
• OK. Kindly notice, $$1-\sin^2\theta-\sin\theta=1\iff -\sin^2\theta-\sin\theta=0$$ multiply both the sides by $-1$, you will get positive sign on RHS as follows $$-1(-\sin^2\theta-\sin\theta)=0\iff \sin^2\theta+\sin\theta=0$$ – Harish Chandra Rajpoot Aug 23 '15 at 3:05