differentiate sin(x+a)=sin(x)cos(a) + cos(x)sin(a) to obtain the addition formula for cosine function?

Answer 1

What we're solving for in this case is cos(x + a), i.e. the addition formula for the cosine function.

Remember that we're doing implicit differentiation with respect to x. Therefore,

sin(x + a) = sin(x)cos(a) + cos(x)sin(a)

Differentiating, keeping in mind that the derivative of a is da/dx, and that we have to use the product rule on the right hand side, we get:

cos(x + a)[1 + da/dx] = cos(x)cos(a) + sin(x)[-sin(a)][da/dx] +
[-sin(x)]sin(a) + cos(x)cos(a)[da/dx]

Expanding out the right hand side, we get

cos(x + a)[1 + da/dx] = cos(x)cos(a) - sin(x)sin(a)[da/dx] - sin(x)sin(a) + cos(x)cos(a)[da/dx]

On the right hand side, let's group the cosines together and the sines together.

cos(x + a)[1 + da/dx] = cos(x)cos(a) + cos(x)cos(a)[da/dx] -
sin(x)sin(a)[da/dx] - sin(x)sin(a)

And now, let's factor the first two terms and the last two terms.

cos(x + a) [1 + da/dx] = cos(x)cos(a) [1 + da/dx] - sin(x)sin(a) [da/dx + 1]

Notice that [da/dx + 1] is a common factor in both grouped terms, so we can use grouping.

cos(x + a) [1 + da/dx] = [1 + da/dx] (cos(x)cos(a) - sin(x)sin(a)]

And now, we can divide both sides by 1 + da/dx

cos(x + a) = cos(x)cos(a) - sin(x)sin(a)

And this is PRECISELY our cosine addition formula.

Answer 2

sin'Y = cos Y.Y'. In this case Y = x + a, so Y' = 1.
(also cos'Y = - sin Y.Y')
d/dx(kf(x)) = kf'(x), where k is a constant.
(sin a & cos a are both constants.)
So differentiating both sides gives us:
cos(x + a) = cos x cos a + (-sin x) sin a
= cos x cos a - sin x sin a
which is the formula for the cosine of a sum.

Answer 3

sin(x+a)=sin(x)cos(a) + cos(x)sin(a)
Differentiating both sides with respect to x (easy, since a is a constant):
cos(x+a) = cos x cos x - sin x sin a

By the way, I'm not sure what you're confused about. The problem tells you what the answer is.