Question

1+sin(пи+альфв )сos(альфа +3пи/2)

Answer 1

$1+sin( \pi +a)cos(a+ \frac{3 \pi }{2})=1-sina*sina=1- sin^{2}a= cos^{2}a$

Answer 2

Let's break down the expression step by step:

1. First, let's simplify the expression inside the sine function:

   π + α = α + π (because addition is commutative) sin(π + α) = sin(α + π) = -sin(α) (because sine is an odd function)

   So, we can replace sin(π + α) with -sin(α) in the expression:

   1 - sin(α) cos(α + 3π/2)

2. Next, let's simplify the expression inside the cosine function:

   α + 3π/2 = 2π - (α/2) (because addition and subtraction of π/2 is equivalent to rotation by 90 degrees) cos(α + 3π/2) = cos(2π - α/2) = cos(α/2) (because cosine is an even function)

   So, we can replace cos(α + 3π/2) with cos(α/2) in the expression:

   1 - sin(α) cos(α/2)

3. We can simplify further by using the double-angle identity for sine:

   sin(2θ) = 2sin(θ)cos(θ)

   Letting θ = α/2, we get:

   sin(α) = 2sin(α/2)cos(α/2)

   Substituting this expression into our original expression gives:

   1 - 2sin(α/2)cos(α/2)cos(α/2)

4. We can simplify further by using the identity:

   cos(2θ) = cos^2(θ) - sin^2(θ)

   Letting θ = α/2, we get:

   cos(α) = cos^2(α/2) - sin^2(α/2)

   Substituting this expression into our previous expression gives:

   1 - 2sin(α/2)(cos^2(α/2) - sin^2(α/2))

   1 - 2sin(α/2)cos^2(α/2) + 2sin^3(α/2)

So the final simplified expression is:

1 - 2sin(α/2)cos^2(α/2) + 2sin^3(α/2)