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Solve in the interval $0^\circ\leq \theta\leq 360^\circ$ the equation $\sin \theta + \cos \theta=1$.

I've got the two angles in the interval to be $0^\circ$ and $90^\circ$, it's not an answer I'm after, I'd just like to see different approaches one could take with a problem like this. Thank you!

Sorry, my approach:

$$\begin{align} \frac{1}{\sqrt 2}\sin \theta + \frac{1}{\sqrt 2}\cos \theta &= \frac{1}{{\sqrt 2 }} \\ \cos 45^\circ\sin \theta + \sin 45^\circ\cos \theta &= \frac{1}{\sqrt 2} \\ \sin(\theta + 45^\circ) &= \frac{1}{\sqrt 2} \\ \theta + 45^\circ &= 45^\circ,\ 135^\circ \\ \theta &= 0^\circ, \ 90^\circ \end{align}$$

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I like your approach best of all. – Lubin Mar 30 '13 at 16:18
@Lubin Thank you – seeker Mar 31 '13 at 11:50
since you are checking the answer in [0,360] so there will be no more values of $\theta$ because sin have +ve in I and II quadrant. – iostream007 May 10 '13 at 13:34
up vote 7 down vote accepted

A slightly 'expanded-upon' version of user67418's answer:

enter image description here

The circle here represents the parametric curve $(x=\cos\theta, y=\sin\theta)$, and the line is the line $x+y=1$, so their points of intersection are the points where $\cos\theta+\sin\theta=1$; at least for me, this is the clearest way of seeing that there are only the two solutions already mentioned.

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Wonderful method. – Thomas Mar 30 '13 at 1:43

I'd write $\sin \theta + \cos \theta = \sqrt{2} \sin \left(\theta + \dfrac{\pi}{4}\right)$ and go from there.

A similar tactic works for all equations of the form $a \sin \theta + b \cos \theta = c$ for constant $a,b,c$.

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Clive i've essentially done this, but could you explain why this works for all equations of that form? I'm finding trouble getting my head around it. How would you know to go from this: sinθ+cosθ to √2sin(θ+π/4)? Is this some sort of trick? – seeker Mar 30 '13 at 0:09
Suppose $$a\cos \theta + b \sin \theta = R\cos(x-\alpha)$$ Don't worry about the fact that I swapped $\sin$ for $\cos$ and $+$ for $-$. Use the angle addition formula to expand the right-hand side of this equation and compare coefficients of $\sin x$ and $\cos x$. You'll see that you get $$R\cos \alpha = a\ \text{and}\ R\sin \alpha = b$$ Solving this then gives $$R^2 = a^2+b^2\ \text{and}\ \tan \alpha = \frac{b}{a}$$ To understand why, either try and work it out (which I recommend), or else I refer you to any good introductory maths textbook. – Clive Newstead Mar 30 '13 at 0:17
Thank you Clive, you've been an enormous help! – seeker Mar 30 '13 at 0:20

Are you familiar with the unit circle? If so, then take that approach to see that $\sin \theta + \cos \theta =1$ is only satisfied when either $\theta = 90^{\circ}$ or $\theta = 0^{\circ}$, or equivalently, $\theta = 360^{\circ}$. This comes from the fact that the unit circle has points $(1,0)$ at $0^{\circ}$, and (0,1) at $90^{\circ}$.

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Nice way to think about it! Thanks! – seeker Mar 30 '13 at 0:11
No problem @Assad; I hope it helps out. – Jamil_V Mar 30 '13 at 0:12

$$1=\sin\theta+\cos\theta\implies\frac{1}{\sqrt 2}=\sin\theta\cos\frac{\pi}{4}+\sin\frac{\pi}{4}\cos\theta=\sin\left(\theta+\frac{\pi}{4}\right)\Longrightarrow$$


If you don't like radians, change

$$\frac{\pi}{4}\sim45^\circ\;,\;\;\frac{3\pi}{4}\sim 135^\circ\;,\;\;\pi\sim 180^\circ$$

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We have $$\sin(\theta) + \cos(\theta) = 1 \text{ and } \sin^2(\theta) + \cos^2(\theta) = 1$$ If we have $$a+b = 1 \text{ and } a^2 + b^2 =1 \tag{$\star$}$$then either $a=0,b=1$ or $a=1, b=0$. This gives us $$\sin(\theta) = 0, \cos(\theta) = 1 \text{ (or) }\sin(\theta) = 1, \cos(\theta) = 0$$Hence, $$\theta = 0^{\circ},90^{\circ}$$

EDIT(Expanding out the implications of $(\star)$).

We shall show here that $(\star) \implies$ either $a=0,b=1$ or $a=1, b=0$.

We have $a=1-b$. Plugging this in $a^2 + b^2 = 1$, we get that $$(1-b)^2 + b^2 = 1 \implies 1-2b+b^2 + b^2 = 1 \implies 2b^2-2b = 0 \implies 2b(b-1) = 0$$ Hence, $b=0$ or $b=1$. Plugging in $b=0$ in $a=1-b$ gives us $a=1$. Similarly, plugging in $b=1$ in $a=1-b$ gives us $a=0$. Hence, the possible solutions are $$(a,b) = (0,1) \text{ or }(1,0)$$

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It might help to put a step in the middle explaining why $a+b=1$ and $a^2+b^2=1$ imply that one variable must be $1$ and the other $0$ (e.g., by computing $(a+b)^2-(a^2+b^2)$). – Steven Stadnicki Mar 30 '13 at 1:02
Thank you @Marvis – seeker Mar 30 '13 at 19:48

Using the fact that $\sin \theta+\cos \theta=\sqrt{2}\cdot\sin (\theta+\frac{\pi}{4})$, this problem is reduced to find an angle $\theta\in [0, 2\pi]$ such that $$\sin (\theta+\frac{\pi}{4})=\frac{\sqrt{2}}{2}.$$ As we know, if $\sin\alpha=\sqrt{2}/2$, it must be the case that $\alpha=\frac{\pi}{4}+2k_1\pi$ or $\alpha=\frac{3\pi}{4}+2k_2\pi$, where $k_1$ and $k_2$ can be any integer. In this problem, since $\theta\in [0,2\pi]$, we have $$\theta+\frac{\pi}{4}\in\bigg[\frac{\pi}{4}, \frac{9\pi}{4}\bigg].$$ Thus,the possible values of $k$'s are: $k_1=0$, $k_1=1$, $k_2=0$, which correspond to three angles in $[0, 2\pi]$: $\theta_1=\frac{\pi}{4}-\frac{\pi}{4}=0$, $\theta_2=\frac{9\pi}{4}-\frac{\pi}{4}=2\pi$, and $\theta_3=\frac{3\pi}{4}-\frac{\pi}{4}=\frac{\pi}{2}$.

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Another approach not so fast in this case but not yet mentioned is a general method to solve any linear equation in $\sin \theta $ and $\cos \theta $ of the form $a\sin \theta +b\cos \theta =c$. We express these functions in terms of $t=\tan \frac{\theta }{2}$: $$ \begin{equation*} \sin \theta =\frac{2t}{1+t^{2}} ,\qquad \cos \theta =\frac{1-t^{2}}{1+t^{2}} \end{equation*}$$ and solve a quadratic equation in $t$. The equation $\sin \theta +\cos \theta =1$ is transformed in $$ \begin{eqnarray*} 2t+1-t^{2}=1+t^{2} &\Leftrightarrow &\left( t-1\right)t=0\Leftrightarrow t=1,t=0. \end{eqnarray*} $$ For $t=\tan \frac{\theta }{2}=1$ we get the solution $\theta =90{{}^\circ}$ and for $t=\tan \frac{\theta }{2}=0$ the solution $\theta=0{{}^\circ}$.

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sin x+cos x=1. Sin^2 x+cos^2 x+2sin(x)cos(x)=1. 2sin(x)cos(x)=0, so either sin(x)=0 or cos(x)=0, giving you the solutions 0, pi/2.

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