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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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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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