Suppose that there is a trigonometric equation of the form $a\sin x + b\cos x = c$, where $a,b,c$ are real and $0 < x < 2\pi$. An example equation would go the following: $\sqrt{3}\sin x + \cos x = 2$ where $0<x<2\pi$.

How do you solve this equation without using the method that moves $b\cos x$ to the right side and squaring left and right sides of the equation?

And how does solving $\sqrt{3}\sin x + \cos x = 2$ equal to solving $\sin (x+ \frac{\pi}{6}) = 1$


The idea is to use the identity $\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$. You have $a\sin x+b\cos x$, so you’d like to find an angle $\beta$ such that $\cos\beta=a$ and $\sin\beta=b$, for then you could write

$$a\sin x+b\cos x=\cos\beta\sin x+\sin\beta\cos x=\sin(x+\beta)\;.$$

The problem is that $\sin\beta$ and $\cos\beta$ must be between $-1$ and $1$, and $a$ and $b$ may not be in that range. Moreover, we know that $\sin^2\beta+\cos^2\beta$ must equal $1$, and there’s certainly no guarantee that $a^2+b^2=1$.

The trick is to scale everything by $\sqrt{a^2+b^2}$. Let $A=\dfrac{a}{\sqrt{a^2+b^2}}$ and $B=\dfrac{b}{\sqrt{a^2+b^2}}$; clearly $A^2+B^2=1$, so there is a unique angle $\beta$ such that $\cos\beta=A$, $\sin\beta=B$, and $0\le\beta<2\pi$. Then

$$\begin{align*} a\sin x+b\cos x&=\sqrt{a^2+b^2}(A\sin x+B\cos x)\\ &=\sqrt{a^2+b^2}(\cos\beta\sin x+\sin\beta\cos x)\\ &=\sqrt{a^2+b^2}\sin(x+\beta)\;. \end{align*}$$

If you originally wanted to solve the equation $a\sin x+b\cos x=c$, you can now reduce it to $$\sqrt{a^2+b^2}\sin(x+\beta)=c\;,$$ or $$\sin(x+\beta)=\frac{c}{\sqrt{a^2+b^2}}\;,$$ where the new constants $\sqrt{a^2+b^2}$ and $\beta$ can be computed from the given constants $a$ and $b$.

  • $\begingroup$ +1 Nice, thoroughly explained answer. I wonder whether there's a trend of some askers to accept the question yet without upvoting it...? $\endgroup$ – DonAntonio Oct 14 '12 at 10:58
  • $\begingroup$ @DonAntonio: Thanks. You need $15$ rep to upvote, so new users don’t have that option. $\endgroup$ – Brian M. Scott Oct 14 '12 at 11:00
  • $\begingroup$ Well, that's a new one for me, @Brian. Thanks for bringing that to my attention. BTW, this seems to me a rather odd rule: if you trust them enough to ask/accept questions trust them to upvote, too. Anyway now I understand better this. $\endgroup$ – DonAntonio Oct 14 '12 at 11:03

Using complex numbers, and setting $z=e^{i\theta}$,

$$a\frac{z-z^{-1}}{2i}+b\frac{z+z^{-1}}2=c,$$ or


The discriminant is $c^2-b^2-a^2:=-d^2$, assumed negative, then the solution

$$z=\frac{c\pm id}{b-ia}.$$

Taking the logarithm, the real part $$\ln\left(\dfrac{\sqrt{c^2+d^2}}{\sqrt{a^2+b^2}}\right)=\ln\left(\dfrac{\sqrt{c^2+a^2+b^2-c^2}}{\sqrt{a^2+b^2}}\right)=\ln(1)$$ vanishes as expected, and the argument is

$$\theta=\pm\arctan\left(\frac dc\right)+\arctan\left(\frac ab\right).$$

The latter formula can be rewritten with a single $\arctan$, using

$$\theta=\arctan\left(\tan(\theta)\right)=\arctan\left(\frac{\pm\dfrac dc+\dfrac ab}{1\mp\dfrac dc\dfrac ab}\right)=\arctan\left(\frac{\pm bd+ac}{bc\mp ad}\right).$$


Riffing on @Yves' "little known" solutions ...

enter image description here

The above trigonograph shows a scenario with $a^2 + b^2 = c^2 + d^2$, for $d \geq 0$, and we see that $$\theta = \operatorname{atan}\frac{a}{b} + \operatorname{atan}\frac{d}{c} \tag{1}$$ (If the "$a$" triangle were taller than the "$b$" triangle, the "$+$" would become "$-$". Effectively, we can take $d$ to be negative to get the "other" solution.) Observe that both $c$ and $d$ are expressible in terms of $a$, $b$, $\theta$: $$\begin{align} a \sin\theta + b \cos\theta &= c \\ b \sin\theta - a\cos\theta &= d \quad\text{(could be negative)} \end{align}$$ Solving that system for $\sin\theta$ and $\cos\theta$ gives $$\left.\begin{align} \sin\theta &= \frac{ac+bd}{a^2+b^2} \\[6pt] \cos\theta &= \frac{bc-ad}{a^2+b^2} \end{align}\quad\right\rbrace\quad\to\quad \tan\theta = \frac{ac+bd}{bc-ad} \tag{2}$$ We can arrive at $(2)$ in a slightly-more-geometric manner by noting $$c d = (a\sin\theta + b \cos\theta)d = c( b\sin\theta - a \cos\theta ) \;\to\; ( b c - a d)\sin\theta = \left( a c + b d \right)\cos\theta \;\to\; (2) $$ where each term in the expanded form of the first equation can be viewed as the area of a rectangular region in the trigonograph. (For instance, $b c \sin\theta$ is the area of the entire figure.)


I'm assuming $ab\neq 0$, since otherwise trivial. $a\sin x+b\cos x-c=0$

$$\iff a\left(2\sin\frac{x}{2}\cos\frac{x}{2}\right)+b\left(\cos^2\frac{x}{2}-\sin^2\frac{x}{2}\right)-c\left(\sin^2\frac{x}{2} +\cos^2\frac{x}{2}\right)=0$$

  • Assume $\cos\frac{x}{2}\neq 0$. Then

$$\stackrel{:\cos^2 \frac{x}{2}\neq 0}\iff (b+c)\tan^2\frac{x}{2}-2a\tan\frac{x}{2}+(c-b)=0$$

If $b+c\neq 0$, then

$$\iff \tan\frac{x}{2}=\frac{a\pm\sqrt{a^2+b^2-c^2}}{b+c}$$

$$\iff x=2\left(\arctan\left(\frac{a\pm\sqrt{a^2+b^2-c^2}}{b+c}\right)+n\pi\right),\, n\in\Bbb Z$$

Real solutions exist iff $a^2+b^2\ge c^2$.

If $b+c=0$, then

$$\iff \tan\frac{x}{2}=\frac{c-b}{2a}\iff x=2\left(\arctan\frac{c-b}{2a}+n\pi\right),\, n\in\Bbb Z$$

  • Assume $\cos\frac{x}{2}=0$. Equality holds iff $b+c=0$.


While the above answers perfectly answer your question, you might benefit from taking a look at the above link. It particularly is relevant for the second part of the question and explains how to simplify $f(x) =a\sin(x) + b\cos(x)$ to a single sine function, approaching it both geometrically and algebraically.

  • $\begingroup$ Linked answers are discouraged, just to let you know, but you did provide some insight on what is found when looking at the link, so good job $(+1)$ $\endgroup$ – Feeds Apr 20 '18 at 3:28

This is little known, but you can solve the equation without much trigonometry.

WLOG, we can assume that $a^2+b^2=1$ (the coefficients can be normalized). Write


The solution of the quadratic equation is $$S=ac\pm bd$$ where $d=\sqrt{1-c^2}$. By symmetry,

$$C=bc\mp ad.$$

If you have enough with the values of the sine and the cosine, you can stop here. Otherwise

$$\theta=\arctan\frac SC.$$

For unnormalized $a,b$, the solution is

$$S=\frac{ac\pm bd}{a^2+b^2},\\ C=\frac{bc\mp ad}{a^2+b^2}$$ and $$\color{green}{\theta=\arctan\frac{ac\pm bd}{bc\mp ad}}$$

where $d=\sqrt{a^2+b^2-c^2}$. It does not exist when $a^2+b^2<c^2$.

From the above, one can observe that the solution is also given by

$$\theta=\arctan\frac ab\pm\arctan\frac dc$$ but this takes two (costly) arc tangents instead of one.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.