Sin(a-3pi/2)cos(4pi+a)

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вопрос

First, we need to simplify the given expression:

sin(a - 3π/2) cos(4π + a)

We know that sin(θ) = cos(π/2 - θ), so we can rewrite sin(a - 3π/2) as cos(3π/2 - (a - 3π/2)) = cos(3π/2 - a + 3π/2) = cos(3π - a)

Similarly, cos(θ) = cos(-θ), so we can rewrite cos(4π + a) as cos(-(4π + a)) = cos(-4π - a) = cos(-a)

Therefore, the expression simplifies to:

cos(3π - a) cos(-a) = cos(3π - a) cos(a)

Now we can use the sum-to-product formula for cosine, which states that cos(A) cos(B) = 1/2 [cos(A - B) + cos(A + B)], to simplify further:

cos(3π - a) cos(a) = 1/2 [cos(3π - a - a) + cos(3π - a + a)]
= 1/2 [cos(2π) + cos(3π)]
= 1/2 [1 + (-1)]
= 1/2 * 0
= 0

Therefore, sin(a - 3π/2) cos(4π + a) simplifies to 0.