Question

Ctg(3p/2+a) tg(p/2-a) -cos(p-a) sin(p/2) -a)

Answer 1

По формулам приведения

сtg(3π/2+α) tg(π/2-α) -cos(π-α) sin(π/2 -α)=-tgα*ctgα-(-cosα)*cosα=

-1+cos²α=-(1-cos²α)=-sin²α

Answer 2

It appears that you have provided an expression involving trigonometric functions, but it seems to be incomplete or contain some errors. Let's try to correct and simplify the expression step by step.

Assuming that you meant to write the expression as:

Ctg(3π/2 + a) * Tg(π/2 - a) - Cos(π - a) * Sin(π/2 - a)

We can simplify it as follows:

1. Cotangent (Ctg) and tangent (Tg) are reciprocal trigonometric functions. Cotangent (Ctg(x)) is equal to 1 divided by tangent (Tan(x)), so Ctg(x) = 1 / Tan(x).

2. The tangent of π/2 - a is the same as the cotangent of a, so Tg(π/2 - a) = Ctg(a).

3. We know that Cos(π - a) is equal to -Cos(a) and Sin(π/2 - a) is equal to Cos(a).

Now, let's simplify the expression:

Ctg(3π/2 + a) * Tg(π/2 - a) - Cos(π - a) * Sin(π/2 - a)

= (1/Tan(3π/2 + a)) * Ctg(a) - (-Cos(a)) * Cos(a)

= Ctg(a) * (1/Tan(3π/2 + a) + Cos^2(a))

You can further simplify it by working with the tangent expression:

1/Tan(3π/2 + a)

The tangent function has a period of π (180 degrees). So, you can add or subtract π from the argument without changing the value. In other words, Tan(x) = Tan(x + π), which means Tan(3π/2 + a) is the same as Tan(π/2 + a).

Now, Tan(π/2 + a) is the tangent of the complementary angle to (π/2 + a), which is (π/2 - a). Therefore:

1/Tan(3π/2 + a) = 1/Tan(π/2 - a) = Ctg(a)

So, the final simplified expression is:

Ctg(a) * (Ctg(a) + Cos^2(a))

Or, you can write it as:

Ctg(a)^2 + Ctg(a) * Cos^2(a)

This is the simplified form of the given expression.