# Solving $\sin \theta + \cos \theta=1$ in the interval $0^\circ\leq \theta\leq 360^\circ$

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}

• I like your approach best of all. Commented Mar 30, 2013 at 16:18
• 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. Commented May 10, 2013 at 13:34

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

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.

• Wonderful method. Commented Mar 30, 2013 at 1:43

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 $$(a,b) = (0,1) \text{ or }(1,0)$$

• 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 $(a+b)^2-(a^2+b^2)$). Commented Mar 30, 2013 at 1:02

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 $a,b,c$.

• 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? Commented Mar 30, 2013 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. Commented Mar 30, 2013 at 0:17

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 $90^{\circ}$.

• Nice way to think about it! Thanks! Commented Mar 30, 2013 at 0:11
• No problem @Assad; I hope it helps out. Commented Mar 30, 2013 at 0:12

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

$$\theta+\frac{\pi}{4}=\begin{cases}\frac{\pi}{4}\\{}\\\pi-\frac{\pi}{4}=\frac{3\pi}{4}\end{cases}\implies\theta=0\;\;\vee\;\;\frac{\pi}{2}$$

If you don't like radians, change

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

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 $\theta=0{{}^\circ}$.

• Ah yes, the tangent half-angle substitution. My favorite way to convert any trigonometric equation into a polynomial. Commented Mar 4, 2018 at 1:56

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 $\theta_3=\frac{3\pi}{4}-\frac{\pi}{4}=\frac{\pi}{2}$.

$\sin x+ \cos x=1 \implies \sin^2 x + \cos^2 x+ 2 \sin x \cos x = 1 \implies 2\sin x \cos x =0$, so either $\sin x=0$ or $\cos x = 0$, giving you the solutions $0$, $\dfrac {\pi}{2}$.