Let f(a) = sin(a) + 1

We need to re-wrtie the function f(a) as a product function of two terms.

Let us use the trigonometric expressions to solve.

We know that:

sin90 = 1

Then we will re-write.

=> f(a) = sina + sin90.

Now we will use the sum of two sines identity.

We know that:

sina + sinb = 2sin(a+b)/2 * cos(a-b)/2

==> sina + sin90 = 2*sin(a+90)/2 * cos(a-90)/2

==>** f(a) = 2sin(a+90)/2 * cos(a-90)/2 **

We have to write sin a + 1 as a product of two terms.

Now 1 = sin 90.

sin a + 1 = sin a + sin 90

Now the sum of two sine terms like sin a + sin b can be written as 2 sin[(a+b)/2] cos[(a-b)/2]

So here 2 sin[(a+b)/2] cos[(a-b)/2]

=> 2* sin [(a + 90)/2] cos[(a - 90)/2]

=> 2* sin (a/2 + 45) * cos ( a/2 - 45)

**Therefore we can write sin a + 1 as 2* sin (a/2 + 45) * cos ( a/2 - 45).**

sina +1 to be shown as a product.

We know that sina = cos(90-a)

We know that cos2A = 2cos^2A -1.

Therefore sina = cos(90-a)

coso(90-a) = 2{cos(90-a)/2)} - 1.

Therefore sina = 2 {cos(90-a)/2}^2 -1.

Therefore sum sina +1 = 2{cos(90-a)/2}^2- 1+1 = 2{cos(90-a)/2}^2 = 2cos(45-a/2)^2, which is in product form.

To change the sum into a product, the given expression has to have 2 like trigonometric functions.

Since we have already the term sin a, we'll write the value 1 as sin pi/2.

We'll re-write the expression:

sin a + sin p/2

To change the sum into a product we'll apply the formula:

sin a + sin b = 2sin[(a+b)/2]cos[(a-b)/2]

We'll let a = a and b = pi/2:

sin a + sin pi/2 = 2sin[(2a+pi)/4]cos[(2a-pi)/4]

sin a + sin pi/2 = 2sin[(a/2 + pi/4)]cos[(a/2 -pi/4)]

**sin a + 1 = 2sin[(a/2 + pi/4)]cos[(a/2 -pi/4)]**