Question

Tgα = -33π/2 < α < 2πНайти:а) ctgαb) 5sinα -2cosα/sinα + 4cosα

Answer 1

Ответ:

Объяснение:

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Answer 2

To solve this problem, let's first find the values of trigonometric functions based on the given value of α.

Given: Tgα = -33π/2, which means -33π/2 < α < 2π

Since tangent is a periodic function with a period of π, we can find an equivalent angle in the interval (0, 2π) by adding or subtracting multiples of π.

Adding 2π to -33π/2 gives us: α = -33π/2 + 2π = -33π/2 + 4π/2 = -33π/2 + 2π/2 = -33π/2 - π/2 = -35π/2

Now we can find the values of trigonometric functions for α:

a) ctgα (cotangent): cot(α) = 1/tan(α) = 1/tan(-35π/2)

To simplify, we need to find the value of tan(-35π/2). Since tangent is periodic with a period of π, we can find an equivalent angle in the interval (0, 2π) by adding or subtracting multiples of π. Adding 4π to -35π/2 gives us:

-35π/2 + 4π = -35π/2 + 8π/2 = -35π/2 + 4π/2 = -35π/2 + 2π = -33π/2

Now we have tan(-35π/2) = tan(-33π/2) = -33π/2 (from the given information).

Therefore, cot(α) = 1/tan(α) = 1/(-33π/2) = -2/(33π).

b) 5sinα - 2cosα / sinα + 4cosα: Let's substitute the value of α into the expression:

5sin(α) - 2cos(α) / sin(α) + 4cos(α) = 5sin(-33π/2) - 2cos(-33π/2) / sin(-33π/2) + 4cos(-33π/2)

Since sine and cosine are periodic functions with a period of 2π, we can find an equivalent angle in the interval (0, 2π) by adding or subtracting multiples of 2π. Adding 4π to -33π/2 gives us:

-33π/2 + 4π = -33π/2 + 8π/2 = -33π/2 + 4π/2 = -33π/2 + 2π = -31π/2

Now we have:

5sin(-33π/2) - 2cos(-33π/2) / sin(-33π/2) + 4cos(-33π/2) = 5sin(-31π/2) - 2cos(-31π/2) / sin(-31π/2) + 4cos(-31π/2)

To simplify further, we need to know the values of sine and cosine at -31π/2. However, this value is not within the standard range for the sine and cosine functions. Please double-check the given range for α or provide additional information if available.