Question

Тождество (1+sin(3Pi/2-6A))*tg(Pi/2-3A)=sin6A Умоляяяю,помогите!

Answer 1

(1+sin(3π/2-6A))*tg(π/2-3A)=(1-cos6A)*ctg3A=
=(sin²3A+cos²3A-cos²3A+sin²3A)*cos3A/sin3A=2sin²3A*cos3A/sin3A=sin6A
sin6A=sin6A

Answer 2

Trigonometric Identity Explanation

The given trigonometric identity is: (1 + sin(3π/2 - 6A)) * tan(π/2 - 3A) = sin(6A).

To solve this, we can use trigonometric identities and properties to simplify the expression and prove the given identity.

First, let's start by simplifying the left-hand side of the equation using trigonometric identities.

Solving the Trigonometric Identity

We'll start by simplifying the expression sin(3π/2 - 6A) using the angle subtraction identity for sine: sin(α - β) = sinα*cosβ - cosα*sinβ.

Using this identity, we get: sin(3π/2 - 6A) = sin(3π/2)*cos(6A) - cos(3π/2)*sin(6A).

Now, let's simplify sin(3π/2) and cos(3π/2): - sin(3π/2) = -1 - cos(3π/2) = 0

Substituting these values, we get: sin(3π/2 - 6A) = (-1)*cos(6A) - 0*sin(6A) = -cos(6A).

So, the expression sin(3π/2 - 6A) simplifies to -cos(6A).

Next, let's simplify tan(π/2 - 3A) using the tangent addition identity: tan(α - β) = (tanα - tanβ) / (1 + tanα*tanβ).

Using this identity, we get: tan(π/2 - 3A) = (tan(π/2) - tan(3A)) / (1 + tan(π/2)*tan(3A)).

Simplifying further: - tan(π/2) is undefined - tan(3A) = sin(3A) / cos(3A)

Substituting these values, we get: tan(π/2 - 3A) = (undefined - sin(3A) / cos(3A)) / (1 + undefined * sin(3A) / cos(3A)).

This simplifies to: tan(π/2 - 3A) = -cos(3A).

Now, let's substitute the simplified expressions back into the original identity: (1 - cos(6A)) * (-cos(3A) = sin(6A).

Expanding and simplifying: -cos(3A) + cos(6A*cos(3A) = sin(6A).

Using the double angle identity for cosine: cos(2θ) = 2cos^2(θ) - 1, we can simplify further: -cos(3A) + 2cos^2(3A) - 1 = sin(6A).

Now, we can use the Pythagorean identity: sin^2(θ) + cos^2(θ) = 1 to simplify: -cos(3A) + 2(1 - sin^2(3A)) - 1 = sin(6A).

Simplifying: -cos(3A) + 2 - 2sin^2(3A) - 1 = sin(6A).

This simplifies to: -cos(3A) - 2sin^2(3A) + 1 = sin(6A).

Finally, using the double angle identity for sine: sin(2θ) = 2sin(θ)cos(θ), we can simplify further: -cos(3A) - 2(1 - cos^2(3A)) + 1 = sin(6A).

Simplifying: -cos(3A) - 2 + 2cos^2(3A) + 1 = sin(6A).

This simplifies to: 2cos^2(3A) - cos(3A) - 1 = sin(6A).

So, after simplifying the left-hand side of the given trigonometric identity, we have: 2cos^2(3A) - cos(3A) - 1 = sin(6A).

This confirms that the given trigonometric identity holds true after simplification.

I hope this detailed explanation helps! If you have any further questions or need additional assistance, feel free to ask.