Question

(1-COSax)/(1-COSbx)
Lim approaches 0, we have to show that that above equation is equal to (a^2)/(b^2)
(1-COSax)/(1-COSbx)
Lim approaches 0, we have to show that that above equation is equal to (a^2)/(b^2)
@Mathematics

Answer 1

you have to use one of these
http://www.sosmath.com/trig/Trig5/trig5/trig5.html

Answer 2

wait as the limit of x apporaches 0?

Answer 3

bt exactly which one?

Answer 4

Multiply top and bottom by (1 + cos ax) and by (1 + cos bx).

Then you'll get an expression with sin^2 and some other terms in cos.

Answer 5

Then you can evaluate the limits in sin^2 by creating terms such as sin(ax)/x, which has limit a.

Answer 6

I.e.,
\[ \frac{1- \cos ax}{1- \cos bx} = \frac{1- \cos^2 ax}{1- \cos^2 bx} \frac{1+ \cos bx}{1+ \cos ax} = \frac{\sin^2 ax / x^2}{\sin^2 bx / x^2} \frac{1+ \cos bx}{1+ \cos ax} \]

Answer 7

y have we divided [\sin ^2 \] by x^2

Answer 8

Because these are now limits we can handle.
\[ \lim_{x \rightarrow 0} \frac{\sin x }{x} = 1 \]
\[ \lim_{x \rightarrow 0} \frac{\sin ax }{x} = \lim_{x \rightarrow 0} \frac{\sin x }{x/a} = a \]

Answer 9

and what we will do (1+cosbx)/(1+cosax)

Answer 10

What is the limit of those terms as x --> 0?

Answer 11

it will become 1 i guess as because cos 0 is 1

Answer 12

Yes, so the limit as x --> 0 of (1+cosbx)/(1+cosax) = (1+1)/(1+1) = 1

Answer 13

thanks alot mate