How do I find the range of $y = a \sin x + b \cos x$

Question

How do I find the range of $$ f(x) = a \sin x + b \cos x $$

My textbook says, $$R_f \equiv \left[ - \sqrt{a^2+ b^2} , \sqrt{a^2 + b^2} \right] $$

But How? Why? How do I prove this?

Answer 1

HINT:

Notice that $\sin^2x+\cos^2x=1\implies \sin x=\sqrt{1-\cos^2x}$ so find the stationary points of the function $$f(x)=a\sqrt{1-\cos^2x}+b\cos x$$

Answer 2

Suppose \begin{align} a\sin x + b\cos x &= R\cos\phi\sin x+R\sin\phi\cos x\\ &= R\sin(x+\phi) \end{align}

Then \begin{align} R\cos\phi&=a\tag{1}\\R\sin\phi&=b\tag{2} \end{align}

$(1)^2+(2)^2$, $$R^2\cos^2\phi+R^2\sin^2\phi=R^2(\cos^2\phi+\sin^2)=R^2=a^2+b^2$$ By convention, $$R=\sqrt{a^2+b^2}$$ but actually it can be negative, it doesn't matter.

So \begin{align} a\sin x + b\cos x &= \sqrt{a^2+b^2}\sin(x+\phi) \end{align}

Since $-1\le\sin x \le 1$, the range is $$\left[-\sqrt{a^2+b^2},\sqrt{a^2+b^2}\right]$$

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To find $\phi$,

$(2)/(1)$, $$\frac{R\sin\phi}{R\cos\phi}=\tan\phi=\frac ba$$ So $$\phi=\arctan \frac ba$$

Answer 3

Let $y=a\sin x+b\cos x$

Using Weierstrass substitution,

we have $$ t^2(b+y)-2at+y-b=0$$ where $t=\tan\dfrac x2$

As $t$ is real, the discriminant must be $\ge0$

i.e., $$(2a)^2\ge4(b+y)(b-y)\iff y^2\le a^2+b^2$$