I have the following function:
$$f(x)=\sqrt{3}\cos (x)+\sin(x)$$
I am trying to find what the amplitude of this function is without graphing it. I understand the formula for a cosine equation is:
$$y=A\cos(Bx-C)$$
Where A is the amplitude, meaning that $\sqrt{3}$ would be the amplitude, but that answer is incorrect. How do you get the amplitude without having to graph?
$\endgroup$4 Answers
$\begingroup$Remember your trig identities.
$$ A\cos(x - C) = A\cos(x)\cos(C) + A\sin(x)\sin(C) $$
So if $f(x) = \sqrt{3}\cos(x) + \sin(x) = A\cos(x-C)$, then $A\cos(C) = \sqrt{3}$ and $A\sin(C) = 1$. If we square both sides and add them together, we get $$ [A\cos(C)]^2 + [A\sin(C)]^2 = A^2[\cos(C)^2 +\sin(C)^2] = A^2 = (\sqrt{3})^2 + (1)^2 = 4 $$ So $A = 2$.
From this argument it should be clear that in the general case, the amplitude of $a\sin(x) + b\cos(x)$ is $\sqrt{a^2 + b^2}$.
$\endgroup$ 2 $\begingroup$The amplitude of $A \cos x + B \sin x$ is $\sqrt{A^2+B^2}$.
You can check easily by differentiating $f(x) = A \cos x + B \sin x$, which gives $\sin x = \frac{B}{\sqrt{A^2+B^2}}$ and $\cos x = \frac{A}{\sqrt{A^2+B^2}}$ and we can get the maximum value of $f(x)$.
So here amplitude is 2.
Hint:
$$\sqrt{3}\cos (x)+\sin(x)=2\biggl(\frac{\sqrt 3}2\cos x+\frac12\sin x\biggr)=\dotsm$$
$\endgroup$ 11 $\begingroup$Use trigonometric identities, you will see that the amplitude is 2. Why? follow these lines
$f(x)=tan(\pi/3)cos(x)+sin(x)=\frac{sin(\pi/3)cos(x)+cos(\pi/3)sin(x)}{cos(\pi/3)}=2sin(x+\pi/3)$
$\endgroup$ 6
