Question

(sin 45+ альфа - cos 45+ альфа)/(sin 45+ альфа +cos 45+альфа) плиз

Answer 1

Sin(45° + α) - cos(45° + α) / sin(45° + α) + cos(45° + α) = (√2/2•cos α + √2/2•sin α - (√2/2•cos α - √2/2•sin α)) / (√2/2•cos α + √2/2•sin α + (√2/2•cos α - √2/2•sin α)) = √2•sin α / √2•cos α = tg α

Answer 2

To simplify the given expression, we'll use the trigonometric identities for the sum of angles:

1. sin(a + b) = sin(a) * cos(b) + cos(a) * sin(b)
2. cos(a + b) = cos(a) * cos(b) - sin(a) * sin(b)

Here, we have sin(45 + α) and cos(45 + α), and we can use the above identities to rewrite them as follows:

sin(45 + α) = sin(45) * cos(α) + cos(45) * sin(α) cos(45 + α) = cos(45) * cos(α) - sin(45) * sin(α)

Since sin(45°) = cos(45°) = 1/√2, and sin(α) is divided by cos(α) in the given expression, the terms containing sin(45 + α) and cos(45 + α) will cancel each other out.

Now, let's simplify the expression:

(sin 45 + α - cos 45 + α) / (sin 45 + α + cos 45 + α)

We get:

(sin 45° * cos α + cos 45° * sin α - cos 45° * cos α + sin 45° * sin α) / (sin 45° * cos α + cos 45° * sin α + cos 45° * cos α - sin 45° * sin α)

Next, group the terms:

[(sin 45° * cos α - cos 45° * cos α) + (cos 45° * sin α + sin 45° * sin α)] / [(sin 45° * cos α + cos 45° * sin α) + (cos 45° * cos α - sin 45° * sin α)]

Now, simplify each group of terms:

[cos α * (sin 45° - cos 45°) + sin α * (cos 45° + sin 45°)] / [cos α * (sin 45° + cos 45°) + sin α * (cos 45° - sin 45°)]

Since sin(45°) = cos(45°) = 1/√2, we can substitute these values:

[cos α * (1/√2 - 1/√2) + sin α * (1/√2 + 1/√2)] / [cos α * (1/√2 + 1/√2) + sin α * (1/√2 - 1/√2)]

Now, simplify further:

[cos α * 0 + sin α * (2/√2)] / [cos α * (2/√2) + sin α * 0]

Finally, we end up with:

(sin α * √2) / (cos α * √2) = sin α / cos α = tan α

So, the simplified expression is tan α.