Simplifying Trig Ratios?

Question

Hey, I need to find out how to simplify a rational expression so I can take the limit as x -->0. Here it is:

sin(2x) / sin(5x)

Thanks!

Answer 1

If you need to know how to do this without using L'Hospital's rule (which would be Calculus I), the only fact you need to know is that

lim (sin(y) / y) = 1
y -> 0

and

lim (y / sin(y)) = 1
y -> 0

Here are the steps to approach this (without L'Hospital's rule).

lim [sin(2x) / sin(5x)]
x -> 0

Multiply the numerator and the denominator by 10x. Normally, we're not allowed to do this, BUT, since we know x is not equal to 0 (by the limit, x is close to but not equal to 0), this is a valid step.

lim [ {10x sin(2x)} / {10x sin(5x)} ]
x -> 0

Now, we can change this into a product.

On the top, decompose 10x into 2 * (5x).
On the bottom, decompose 10x into 5 * (2x).

lim [ {2*(5x) sin(2x)} / { 5*(2x) sin(5x)} ]
x -> 0

Pull the constant 2 on the top out of the limit, and the constant 5 on the bottom out of the limit. Done together, this puts a 2/5 on the outside.

(2/5) lim [ { (5x) sin(2x) } / {(2x) sin(5x)} ]
. . . . x -> 0

Now, change the fraction to a product of fractions in this way:

(2/5) lim [ { (5x) / sin(5x) } / {sin(2x) / (2x)} ]
. . . . x -> 0

Remember that the limit of a product is the product of limits.

(2/5) lim [(5x) / sin(5x)] lim [sin(2x) / (2x)]
. . . . . x -> 0 . . . . . . . . . . x -> 0

And now, as per the fact that lim (sin(y)/y) = 1 as y -> 0, and
lim (y/siny) = 1 as y -> 0, both of those limits (which are now in that form) are equal to 1, so we have

(2/5) (1) (1)

Which is equal to 2/5.

Answer 2

It's 2/5. This follows from the expansion of sin z, where z is any angle expressed in radians. (Actually, since you're taking the RATIO of two sines as the argument tends to zero, you can express it in any damned measure of angle that you please, including radians, degrees, or even squiggleyflops, the measure preferred by we Vulcans!)

In radians,

sin z = z - z^3 / 3! + z^5 / 5! - z^7 / 7! + ... (well, you get the picture!)

= z (1 - z^2 / 3! + z^4 / 5! - z^6 / 7! + ... ).

It's clear that in the limit, as z ---> 0, (sin z) / z ---> 1 (all the "correction terms" in z^2 and higher powers clearly ---> 0), that is sin z ---> z.

So [sin (2x)] / [sin (5x)] ---> [2x] / [5x] = 2/5. QED.

Live long and prosper.

POSTSCRIPT : Steve's 2nd equation just HAS to be WRONG! He writes:

Sin 5x = 5sinx -20sin³x + 5cosx.

But, as x ---> 0, sin 5x ---> 0, sin x ---> 0, and cos x ---> 1; so his equation says, in the limit :

0 = 0 - 0 + 5.

'Nuff said !

Answer 3

This may or may not help:
Sin 2x = 2sinx*cosx
Sin 5x = 5sinx -20sin³x + 5cosx