Can someone please help me solve this?
$$\cos^{2}{\theta}-\sin{\theta} = 1, \quad\theta\in[0^{\circ}, 360^{\circ}]$$
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Harish Chandra Rajpoot
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barryedx
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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 02:44
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2Hint: $\sin^2\theta + \cos^2\theta = 1$. – vadim123 Aug 23 '15 at 02:45
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Is that $360^\circ$ or $360$ radians? – Rocket Man Aug 23 '15 at 02:50
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It is 360 degree AJ Stas – barryedx Aug 23 '15 at 02:52
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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 02:54
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@barryedx : Click the 'edit' button in any post to find out how to do the Latex. For theta, use \theta surrounded by dollar signs ( and you know it is always about dollars ;). – Gary. Aug 23 '15 at 03:56
2 Answers
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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 $$\color{red}{\theta}=\left\{\color{blue}{0^\circ, 180^\circ, 270^\circ, 360^\circ} \right\}$$
Harish Chandra Rajpoot
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Please explain how the negative sign turn to positive in step no. 3, thanks – barryedx Aug 23 '15 at 02:59
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5OK. 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 03:05
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$\cos^{2}(\theta) - \sin(\theta) = \cos^{2}(\theta) + \sin^{2}(\theta)$, (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
$0 = \sin(\theta) \Rightarrow \theta = 0^{\circ},180^{\circ},360^{\circ}$
$-1 = \sin(\theta) \Rightarrow \theta = -270^{\circ}$
$\theta = \{0^{\circ},180^{\circ},270^{\circ},360^{\circ}\}$.
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4You really don't need the quadratic formula to solve the equation $x^2 + x = 0$. – anomaly Aug 23 '15 at 03:38
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1@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 03:56
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1Come 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 05:03