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Suppose that there is a trignometric 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 $\sin (x+ \frac{\pi}{6}) = 1$

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Hint: Expand $\sin(x+\frac{\pi}{6})$. –  ᴊ ᴀ s ᴏ ɴ Oct 14 '12 at 9:42
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I explained the general solution in my answer here: math.stackexchange.com/a/201405/4583 –  Ayman Hourieh Oct 14 '12 at 9:51
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1 Answer

up vote 6 down vote accepted

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 $b$.

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+1 Nice, thoroughly explained answer. I wonder whether there's a trend of some askers to accept the question yet without upvoting it...? –  DonAntonio Oct 14 '12 at 10:58
    
@DonAntonio: Thanks. You need $15$ rep to upvote, so new users don’t have that option. –  Brian M. Scott Oct 14 '12 at 11:00
    
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. –  DonAntonio Oct 14 '12 at 11:03
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