# Does $\sin (5x)=5\sin(x)$? Why or why not?

This may sound like a dumb question, but does $\sin(5x)=5\sin(x)$? Can you say this for any trigonometric function if it is true (e.g. can $\tan^{-1}(7x)=7\tan^{-1}(x)$). Again, I am pretty sure I am wrong, but I am just wondering for my future work on trigonometry. If they are not equal, then please tell me what the difference is between the two. Thanks!

Note: I have not had much practice with trigonometry, so please do not answer with some kind of complicated Taylor Series thing (I have heard about it, but don't fully understand it). Thanks again

• They are not equal. A most obvious reason is that $\sin(5x)$ never exceeds $1$ (being equal to the $y$-coordinate of a point on the unit circle). But $5\sin x$ can be as large as $5$ (being equal to the $y$-coordinate of a point on the unit circle multiplied by five). – Jyrki Lahtonen Feb 26 '14 at 6:16
• The $\sin$ function is NOT linear. – IAmNoOne Feb 26 '14 at 6:21
• Check this out, it could be helpful! :) – mikhailcazi Feb 26 '14 at 6:51
• There should be a few values $x$ for which you are familiar with the value $\sin (x)$. (If there aren't, you should learn a few! Having a few examples/computations to keep in mind when thinking about mathematics is vital. I cannot stress this enough.) Does your conjecture hold for these values? – Thomas Belulovich Feb 27 '14 at 2:38
• I should have thought of this before I asked the question.... $\sin(3\times 30)=1$, but $3\sin(30)=3\times 0.5=1.5$! – TrueDefault Feb 27 '14 at 2:51

I'll try to explain the difference between the two with a picture: (Apologies ahead; it's a rough sketch done in Microsoft Paint.)

Here, we have a circle (suppose with radius 1) with its center at (0,0), with a line (call it $l$) drawn from the center to the edge of the circle (ending at $P$). $A$ is the angle the line makes with the $x$ axis.

To get $5\sin(A)$, we blow up the line and the circle to 5 times its original size. $l$ is now 5 times the length, but still has one point at the center of the circle, and makes the same angle with the $x$ axis. The $y$-coordinate of the other end of $l$ is $5\sin(A)$.

To get $\sin(5A)$, this time we don't blow the circle and the line bigger. Instead, we're rotating the line around the center of the circle so that the angle it makes with the $x$-axis is 5 times bigger. Now, the $y$-coordinate of the point where $l$ meets the edge of the circle is $\sin(5A)$.

Try it out with an example or two; you'll see the difference yourself.

• I think your answer does the best job at helping the OP: thank you for the effort. – Eric Stucky Feb 26 '14 at 6:59
• @EricStucky Yeah, I figured OP was looking for an explanation in simpler terms. So I provided one. – Dennis Meng Feb 26 '14 at 7:01

No, placing a constant factor inside of the sine function, as in sin(a*x) changes the frequency of the function; if |a| is greater than one, it "squishes" the sine function horizontally, making it oscillate "faster", and if |a| is less than one, it "stretches out" the function horizontally, making it oscillate "slower". Placing the factor in front of the function, as in a*sin(x), scales the amplitude of the function, "stretching", or "squishing" it vertically.

The best way to observe it is to look at a graph and to see what it does: http://www.wolframalpha.com/input/?i=plot+sin+x%2C+sin+3x%2C+3*sin+x

I suggest playing around with other values in the graphs to see what they do.

1) You could have just plotted the functions in, e.g., wolfram alpha and see for yourself.

2) for $t=\frac{\pi}{2}$ one has $\sin (t)=1$ and since $|\sin (x)|\le 1$, it is clear that $\sin (5\cdot t)\ne 5\cdot \sin(t)$.

Something like $a\cdot f(x)= f(a\cdot x)$ is true if and only if $f(x)$ is linear. Since there are always nonlinear terms in the Taylor expansion of a trigonometric function this can't be true for any.

• Isn't $f(x)=x+1$ linear? But $5f(x)=5x+5$ and $f(5x)=5x+1$. – Joel Reyes Noche Feb 26 '14 at 6:37
• Functions satisfying the stated property are called homogenous, not linear. Any linear function is homogenous, but not necessarily vice versa. As for $x\mapsto x+1$, it is not a linear function. Rather it is an example of an affice mapping. – Ittay Weiss Feb 26 '14 at 6:49
• the argument that since the taylor expansion contains nonlinear terms, then the function itself is not linear is true, but requires some proof (easy). But, for the question posed resorting to the Taylor expansion of the function is really an overkill. Simply plugging in some values into the purported equation immediately shows it does not hold. – Ittay Weiss Feb 26 '14 at 6:51
• I was assuming that the OP (not the answerer) was more familiar with calculus rather than with linear algebra. See Wikipedia, where a linear function is sometimes defined as a polynomial function of degree zero or one. – Joel Reyes Noche Feb 26 '14 at 7:02

Assume that the identity $\sin(5x) = 5\sin(x)$ is true. Since we know that $\sin(180^\circ) = 0$, we'll have: $0 = \sin(180^\circ) = \sin(5*36^\circ) = 5*\sin(36^\circ)$, hence $5*\sin(36^\circ) = 0$, which mounts to $\sin(36^\circ) = 0$. Is this true?

• Using degrees rather than radians should be discouraged. – Ittay Weiss Feb 26 '14 at 6:52

The trigonometric functions are functions and as such are usually represented using function notation. Recall that the function notation $f(x)$ means that function $f$ is accepting $x$ as its input. It does not mean that $f$ and $x$ are being multiplied. So the notation $\sin(5x)$ means that the sine function is accepting $5x$ as its input. The notation $5\sin(x)$ means that 5 is being multiplied to the output of the sine function when its input is $x$. Is everything clear now?

• Just to clarify that there exist functions with the property that $f(c\cdot x)=c\cdot f(x)$. Such functions are called homogenous. – Ittay Weiss Feb 26 '14 at 6:19

Both of the equalities that you wrote above are NOT true. One way to look at this might be to use their Taylor series expansion. $\sin (5x)$ means the sine of $5$ times an angle measure $x$. This means in a right triangle whose hypotenuse is of length $1$, the perpendicular corresponding to the acute angle $5x$ has length $\sin (5x)$.

While for $5 \sin x$, one should consider a right-triangle with hypotenuse of length one and the perpendicular corresponding to an acute angle $x$ and then take $5$ times the length of this perpendicular.

You can draw some such right triangles to verify the results yourself.

• Using Taylor series here is a major overkill. Simply plug in some values and observe. – Ittay Weiss Feb 26 '14 at 6:20
• @IttayWeiss haha. I can only agree :-) – Singhal Feb 26 '14 at 7:44