Graphical Interpretation of the Trigonometric identity $A\cos(\theta)+B\sin(\theta)$

Question

I have been taught how to manipulate any expression given as

$$A\cos(\theta)+B\sin(\theta)$$

and then getting useful results by writing it as

$$\sqrt{A^2+B^2}\left(\frac{A\cos(\theta)}{\sqrt{A^2+B^2}}+\frac{B\sin(\theta)}{\sqrt{A^2+B^2}}\right)$$

Now we just made an right angled Triangle with sides $$A,B,\sqrt{A^2+B^2}$$ Now in this triangle one angle was taken as $\phi$..and so the expression above was rewritten as

$$(\sqrt{A^2+B^2})(\sin(\phi)\cos(\theta)+\sin(\theta)\cos(\phi))$$ or popularly as $$\sqrt{A^2+B^2}(\sin(\theta+\phi))$$... while this is very clear to me but I haven't been able to get the idea of $\phi$ in the question ..it wasn't even in the question after all...it was like we just made it up!!..I think the triangle means something on the graph of sine or cosine or something like that.. so.does anybody know the graphical or intuitive or physical interpretation of the triangle and angle $\phi$ ??it must be having some special meaning..in the graph..atleast intuitively..

Answer 1

The basic intuition that I apply to this question is that we are dealing here with various sinusoidal functions with period $2\pi$. The function $\theta \mapsto \sin(\theta)$ is one such function, and $\theta \mapsto \cos(\theta)$ is another such function.

In fact, every sinusoidal function of $\theta$ with period $2\pi$ is either $\sin(\theta)$ or some stretched-and/or-translated version of $\sin(\theta)$, such as $2 \sin(\theta)$ (whose graph has peaks and valleys twice as far from the $\theta$-axis as the graph of $\sin(\theta)$) or $\sin\left(\theta - 1\right)$ (whose graph looks exactly like the graph of $\sin(\theta)$ except that it is shifted one unit to the right).

The cosine function also is a sinusoidal function with period $2\pi$:

$$\cos(\theta) = \sin\left(\theta + \frac\pi2\right).$$

The fact is that if you take any linear combination of sinusoidal functions with period $2\pi$, you get a new function that also happens to be a sinusoidal function. For example, consider the function $$f(\theta) = \frac12 \sin(\theta) - \frac32 \cos(\theta).$$

As we can see from the graph of this function, the result of combining the sine and cosine functions in this way is a graph that looks just like the graph of the sine function, but with higher peaks and shifted slightly to the right.

If we want to talk about a linear combination of sine and cosine in the most general way, we can write

$$g(\theta) = A\cos(\theta)+B\sin(\theta)$$

which says that $g(\theta)$ is a function formed by taking some linear combination of sine and cosine, literally any linear combination that you like. (Just pick any real-numbered values to plug in for $A$ and $B$!) The fact that

$$A\cos(\theta)+B\sin(\theta) = \sqrt{A^2+B^2}(\sin(\theta+\phi))$$

is a fact that is proved via the manipulations you have shown. That is, these manipulations show that the function $g(x)$ is basically a sine function, except that it is stretched vertically by a factor of $\sqrt{A^2+B^2}$ and it has been shifted to the left by $\phi$ units (or to the right by $-\phi$ units; those are two ways of describing the same movement).

In other words, we prove that $\theta \mapsto A\cos(\theta)+B\sin(\theta)$ is a sinusoidal function by finding out how much we have to stretch and shift the sine function in order to get something equal to $A\cos(\theta)+B\sin(\theta)$. And $\phi$ is just a symbol that we use to mean "how far we have to shift the function".

Answer 2

In fact, $\phi$ is not arbitrary! You should find a $\phi$ in the interval $[0,2 \pi)$ for given values of $A$ and $B$ such that

$$\eqalign{ & \sin \phi = {A \over {\sqrt {{A^2} + {B^2}} }} \cr & \cos \phi = {B \over {\sqrt {{A^2} + {B^2}} }} \cr} $$

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Example

Consider $A={1 \over \sqrt3}$ and $B=1$. Then you get

$$\left\{ \matrix{ \sin \phi = {{{1 \over {\sqrt 3 }}} \over {\sqrt {{1 \over 3} + 1} }} = {1 \over 2} \hfill \cr \cos \phi = {1 \over {\sqrt {{1 \over 3} + 1} }} = {{\sqrt 3 } \over 2} \hfill \cr} \right.\,\,\,\,\, \to \,\,\,\,\,\phi = {\pi \over 6}$$

Here is an image comparing the graph of $\sqrt {{A^2} + {B^2}} \sin \left( {\theta + \phi } \right)$ in $\color{green}{green}$ with the graph of $\sqrt {{A^2} + {B^2}} \sin \left( \theta \right)$ in $\color{blue}{blue}$. The length of the $\color{red}{red}$ line equals the angle $\phi$ in radian.

Answer 3

This is coming from the sin of sum formula.

A nice geometric proof of this is given by Nelson in "Proofs without words", see picture below:

To understand it, construct the following picture around your triangle:

In the first picture above replace the pink triangle by your triangle with sides $A,B, \sin{A+B}$ and draw two triangles with angle $\alpha=\theta$.

Now calculate the lengths of the rectangle in terms of $A,B, \theta$, and then look at the white triangle.

Added about the graphical part This can be done easily but you need to work with complex numbers, and then you lose insight as graphs of complex functions cannot be drawn to well: Consider $A,B \in \mathbb R$.

$$A\cos(\theta)+B\sin(\theta)= \frac{A-Bi}{2} [\cos(\theta)+i \sin(\theta)]+ \frac{A+Bi}{2} [\cos(\theta)-i \sin(\theta)]$$

Graphically, $\frac{A-Bi}{2} [\cos(\theta)+i \sin(\theta)]$ rotates the complex number $\frac{A-Bi}{2} $ by an angle of $\theta$ and $\frac{A+Bi}{2} [\cos(\theta)-i \sin(\theta)]$ rotates the number $\frac{A+Bi}{2}$ by $-theta$.

To get the formula, all you need to see is that if you start from the complex number $z =A+Bi$, you rotate $\frac{\bar{z}}{2}$ by $\theta$, rotate $\frac{z}{2}$ by $-\theta$, and add them together, you get the imaginary part of ( $z$ rotated by $\theta$).