Question

Sin(п-a) sin(2п+a)/tg(п+a) cos(п/2+a)=-cosa

Answer 1

Вместо ( х ) напиши альфа)

Answer 2

sinx*sinx/tgx*(-sinx)

sinx/-tgx

sin*(-cosx/sinx)

-cosx=-cosx

Отже, тотожність доведена

Answer 3

Let's prove the trigonometric identity step by step:

1. Recall some basic trigonometric identities:
   • sin(π - θ) = sin(θ)
   • sin(2π + θ) = sin(θ)
   • tan(π + θ) = -tan(θ)
   • cos(π/2 + θ) = -sin(θ)

Now, let's start with the left-hand side (LHS) of the given identity:

LHS = sin(π - a) * sin(2π + a) / tan(π + a) * cos(π/2 + a)

2. Applying the first two trigonometric identities:

LHS = sin(a) * sin(a) / (-tan(a)) * (-sin(a))

3. Using the fact that tan(θ) = sin(θ) / cos(θ):

LHS = sin(a) * sin(a) / (sin(a) / cos(a)) * (-sin(a))

4. Canceling out sin(a) in the numerator and denominator:

LHS = sin(a) / (1 / cos(a)) * (-sin(a))

5. Simplifying the expression on the left-hand side:

LHS = sin(a) * cos(a) * (-sin(a))

6. Rearranging the terms:

LHS = -sin(a) * sin(a) * cos(a)

7. Using the identity sin^2(θ) = 1 - cos^2(θ):

LHS = -(1 - cos^2(a)) * cos(a)

8. Expanding the expression:

LHS = -cos(a) + cos^3(a)

Now, let's examine the right-hand side (RHS) of the given identity:

RHS = -cos(a)

Since the left-hand side (LHS) is equal to -(1 - cos^2(a)) * cos(a) and the right-hand side (RHS) is equal to -cos(a), we have:

LHS = RHS

Therefore, the given trigonometric identity is proven:

sin(π - a) * sin(2π + a) / tan(π + a) * cos(π/2 + a) = -cos(a)