For the given : $f(t) = 2sint + \sqrt{3}cost$, we are tasked to find the amplitude, period and phase shift.
From the book :
- $f(t) = 2sint + \sqrt{3}cost$ = a sin (t + c)
- $a = \sqrt{2^2 + (\sqrt{3})^2} = 4$
- $tan c = \frac{\sqrt{3}}{2} $ ; $ c = tan^{-1}({\frac{\sqrt{3}}{2}}) = \frac{\pi}{3}$
- $f(t) = 2sint + \sqrt{3}cost$ = $4 sin (t + \frac{\pi}{3})$
Final Answer : $ f(t) =4 sin (t + \frac{\pi}{3})$ where A = 4, period = $2*\pi$ and phase shift = $\frac{\pi}{3} $ to the left.
My questions :
- On the first bullet, how did he know that the given function is equal to $a sin(t + c)$ by looking at the equation f(t) = sin + cos?
- On the 3rd bullet, $tanc = \frac{\sqrt{3}}{2}$; Why is the $\sqrt{3}$ on the numerator and 2 on the denominator? Tan is equal to $\frac{sin}{cos}$, so from the given $f(t) = 2sint + \sqrt{3}cost$, why is the $\sqrt{3}$ the value for sine (numerator) when it is in the term containing cosine?
- On bullet 3 & 4, why is the value of c = $\frac{\pi}{3}$? I mean, yeah the c is solved via the preceding formula but why is that the formula for the c? The value that will determine the phase shift of the function?
Note :
The answer on the book maybe wrong as the amplitude must be equal to $\sqrt{7}$ and $c = tan^{-1}({\frac{\sqrt{3}}{2}})$ is not equal to $ \frac{\pi}{3}$. But even though the answer is wrong, the method of solving is right. (I think.. based on the other problems.)
Again, thank you in advance. Any help would be appreciated.
$\endgroup$ 31 Answer
$\begingroup$This is achieved through the linear combination of $\sin$ and $\cos$:
$$a\sin x + b\cos x = c\sin(x + \varphi)$$
Where:
$$c = \sqrt{a^2 + b^2}$$ $$\varphi = \text{atan2}(b, a)$$
In this case $\text{atan2}(b, a)$ represents a generalized $\arctan(b/a)$. Thus:
$$c = \sqrt{2^2 + \sqrt{3}^2} = \sqrt 7\neq 4$$ $$\varphi = \arctan \frac {\sqrt3}2 \implies \tan \varphi = \frac {\sqrt3} 2$$
Notice that $\arctan(\sqrt 3/2)$ is not $\arctan \sqrt 3$ so it is not $\pi/3$. It is approximately $0.714 \text{ rad}$:
$$2\sin t + \sqrt 3 \cos t \approx \sqrt 7 \sin(t + 0.714)$$
I think you'll need to get a new book if it's making such blatant errors.
$\endgroup$ 4
