Question

an hour passed and i'm still stuck in this qn ):

Answer 1

cos pi/6= ?? @neuronsmisfiring

Answer 2

(sqrt3)/2?

Answer 3

-1/2

Answer 4

use cos a - cos b fomula and just simplfy it

Answer 5

you mean this one? cos(A - B) = cos A cos B + sin A sin B

Answer 6

It's a 0/0

Answer 7

nope cos A - cos B

Answer 8

I think L'hopitals will apply?

Answer 9

-(lim x->pi/6 sin(x) ) = -1/2

Yes apply L'hopitals.

Answer 10

@hba how do u get -1/2? pls explain

Answer 11

if you dont want to use L'hopital, try below for numerator, and maybe you have to group sinx/x..

CosC - CosD = 2Sin[(C+D)/2]Sin[(D-C)/2]

Answer 12

Okay let me solve the whole thing then,Apply L'Hospital's rule.
\[\lim_{\theta \rightarrow \pi/6} \frac{d/d \theta( -\sqrt{3}/2+\cos \theta) }{ d/d \theta(-\pi/6+\theta) }\]

Answer 13

@neuronsmisfiring , you must derive the numerator and the denominator seperately and then substitute. With that you will arrive at the limit.

Answer 14

so u mean not applying quotient rule or applying? @Yttrium

Answer 15

Not applying.

Answer 16

so -sin(theta)/1?

Answer 17

Yep.

Answer 18

hm, i didnt know can differentiate separately.. omg

Answer 19

Thank you! :D doubts cleared!

Answer 20

No problem. :)) Btw, you can apply the rule as many as you can as long as you arrive at an answer that is not indeterminate. :))

Answer 21

okay thanks ytty! :D