Question

16. Find limt
sin (a +3h)-3 sin(a+2h)+3 sin (a+h)-sin a/h cube​

Answer 1

What is the value of lim [sin (a+3h)-3sin (a+2h) +3sin(a+h)-sina] /h raised to 3 as h tends to 0?

Let x=

limh→0sin(a+3h)−3sin(a+2h)+3sin(a+h)−sinah3

In the numerator,

sin(a+3h)−sina=2⋅sin(a+3h−a2)⋅cos(a+3h+a2)

=2sin(3h2)cos(2a+3h2)

And,

−3sin(a+2h)+3sin(a+h)=3⋅2⋅sin(a+h−a−2h2)⋅cos(a+2h+a+h2)

=−6sin(h2)cos(2a+3h2)

Therefore, the numerator becomes

2sin(3h2)cos(2a+3h2)−6sin(h2)cos(2a+3h2)

=2cos(2a+3h2)[sin(3h2)−3sin(h2)]

Now, using triple angle identity for sin(3h2) ,

=2cos(2a+3h2)[3sin(h2)−4sin3(h2)−3sin(h2)]

=−8cos(2a+3h2)sin3(h2)

Therefore,

x=limh→0−8cos(2a+3h2)⋅[sin(h2)h2]3⋅18

Applying limits by using:

limθ→0sinθθ=1

We get,

x=−cos(2a+3⋅02)

x=−cosa

Hence,

limh→0sin(a+3h)−3sin(a+2h)+3sin(a+h)−sinah3

=−cosa

Hope that helps :)

Answer 2

Answer:

Adi pls help me.

I need your help.