Question

Sin^2 альфа + sin(пи-альфа) cos(пи/2-альфа)/tg(пи+альфа)ctg(3пи/2-альфа)

Answer 1

Здесь используются формулы приведения (об этом можно почитать в интернете).

$\sin^2\alpha +\dfrac{\sin(\pi -\alpha)\cos\left(\frac{\pi}{2}-\alpha\right)}{{\rm tg}\left(\pi+\alpha\right){\rm ctg}\left(\frac{3\pi}{2}-\alpha\right)}=\sin^2\alpha+\dfrac{\sin \alpha\cdot\sin\alpha}{{\rm tg}\alpha\cdot{\rm tg}\alpha}=\sin^2\alpha+\dfrac{\sin^2\alpha}{{\rm tg}^2\alpha}=\\ \\ \\ =\sin^2\alpha+\dfrac{\sin^2\alpha}{\dfrac{\sin^2\alpha}{\cos^2\alpha}}=\sin^2\alpha+\cos^2\alpha=1$

Answer 2

Let's break down the expression step by step:

1. sin^2(α): This represents the square of the sine of angle α.

2. sin(π - α): This is the sine of the angle (π - α), which is equal to the sine of α. (sin(π - θ) = sin(θ))

3. cos(π/2 - α): This is the cosine of the angle (π/2 - α), which is equal to the sine of α. (cos(π/2 - θ) = sin(θ))

4. tg(π + α): This represents the tangent of the angle (π + α).

5. ctg(3π/2 - α): This represents the cotangent of the angle (3π/2 - α), which is equal to the tangent of α. (ctg(3π/2 - θ) = tan(θ))

Now, let's substitute these values into the expression:

sin^2(α) + sin(π - α) * cos(π/2 - α) / tg(π + α) * ctg(3π/2 - α)

= sin^2(α) + sin(α) * sin(α) / tg(π + α) * ctg(3π/2 - α)

= sin^2(α) + sin^2(α) / tg(π + α) * ctg(3π/2 - α)

= sin^2(α) + sin^2(α) / [tan(π + α) * tan(3π/2 - α)]

= sin^2(α) + sin^2(α) / [-tan(α) * tan(α)]

= sin^2(α) - sin^2(α)

= 0

Therefore, the expression simplifies to 0.