Question

Cos3a*cos2a-sin2a*sin3a при a= -p/5​

Answer 1

cos(3a)•cos(2a) - sin(2a)•sin(3a) = cos(3a+2a) = cos(5a) = cos(-π) = cos(π) = -1

Answer 2

To evaluate the expression cos(3a) * cos(2a) - sin(2a) * sin(3a) when a = -π/5, we can substitute the value of a into the trigonometric functions and simplify.

Let's start by evaluating cos(3a) and sin(3a) when a = -π/5:

cos(3a) = cos(3 * (-π/5)) = cos(-3π/5) = cos(3π/5) [cos(-θ) = cos(θ)] = -cos(π/5) [cos(θ) = cos(-θ)] = -cos(π/5)

sin(3a) = sin(3 * (-π/5)) = sin(-3π/5) = -sin(3π/5) [sin(-θ) = -sin(θ)] = -sin(π/5) [sin(θ) = sin(-θ)]

Next, let's evaluate cos(2a) and sin(2a) when a = -π/5:

cos(2a) = cos(2 * (-π/5)) = cos(-2π/5) = cos(2π/5) [cos(-θ) = cos(θ)] = cos(π/5)

sin(2a) = sin(2 * (-π/5)) = sin(-2π/5) = -sin(2π/5) [sin(-θ) = -sin(θ)] = -sin(π/5) [sin(θ) = sin(-θ)]

Now, substitute these values into the expression cos(3a) * cos(2a) - sin(2a) * sin(3a):

cos(3a) * cos(2a) - sin(2a) * sin(3a) = (-cos(π/5)) * (cos(π/5)) - (-sin(π/5)) * (-sin(π/5)) = -cos^2(π/5) + sin^2(π/5)

Since cos^2(θ) + sin^2(θ) = 1 (the Pythagorean identity), we can simplify further:

-cos^2(π/5) + sin^2(π/5) = -1 + sin^2(π/5) = sin^2(π/5) - 1

Therefore, when a = -π/5, the value of the expression cos(3a) * cos(2a) - sin(2a) * sin(3a) is sin^2(π/5) - 1.