Question

Cds4альфа+1=1/2sin4альфа(ctgальфа-tgальфа)

Answer 1

$cos^22a-sin^22a+cos^22a+sin^22a=2sin2a*cos2a( \frac{cos^2a-sin^2a}{2sina*cosa} ) \\2 cos^22a=2sin2a*cos2a \frac{cos2a}{sin2a} \\2 cos^22a=2cos^22a$
Получили верное равенство. Однако нельзя забывать, что тангенс и котангенс определены не при всех х. Значит  $x \neq \frac{ \pi n}{2}$ , n∈Z
Решениями будут являться промежутки $(\frac{ \pi n}{2};\frac{ \pi (n+1)}{2})$

Answer 2

To solve the equation you provided, we'll simplify and manipulate the terms step by step. Let's start:

Cds4α + 1 = (1/2)sin(4α)(ctgα - tgα)

First, let's simplify the trigonometric identities:

ctgα = 1/tanα tgα = sinα/cosα

Now we can substitute these values back into the equation:

Cds4α + 1 = (1/2)sin(4α)((1/tanα) - (sinα/cosα))

Next, let's simplify the expression further:

Cds4α + 1 = (1/2)sin(4α)((cosα - sinα)/cosα)

We can expand the sin(4α) term using the double-angle formula:

Cds4α + 1 = (1/2)(2sin(2α)cos(2α))((cosα - sinα)/cosα)

Simplifying further:

Cds4α + 1 = sin(2α)cos(2α)(cosα - sinα)/cosα

We can rewrite sin(2α) as 2sinαcosα:

Cds4α + 1 = 2sinαcosαcos(2α)(cosα - sinα)/cosα

Now, we have cos(2α) = cos²α - sin²α:

Cds4α + 1 = 2sinαcosα(cos²α - sin²α)(cosα - sinα)/cosα

Expanding further:

Cds4α + 1 = 2sinαcosα(cos³α - sin²αcosα - cosαsin²α + sin³α)/cosα

Now, we can simplify sin²αcosα and cosαsin²α as sinαcosα:

Cds4α + 1 = 2sinαcosα(cos³α - 2sinαcosα + sin³α)/cosα

Next, let's simplify sin³α and cos³α using the identities:

sin³α = (1 - cos²α)sinα = sinα - cos²αsinα cos³α = (1 - sin²α)cosα = cosα - sin²αcosα

Substituting these back into the equation:

Cds4α + 1 = 2sinαcosα((sinα - cos²αsinα) - 2sinαcosα + (cosα - sin²αcosα))/cosα

Simplifying further:

Cds4α + 1 = 2sinαcosα(sinα - cos²αsinα - 2sinαcosα + cosα - sin²αcosα)/cosα

Combining like terms:

Cds4α + 1 = 2sinαcosα(sinα - 2sinαcosα + cosα - cos²αsinα - sin²αcosα)/cosα

Now, let's distribute sinα and cosα:

Cds4α + 1 = 2(sin²α - 2sin²αcosα + cosαsinα - cos²αsinα - sin²αcosα)/cosα

Combining like terms again:

Cds4α + 1 = 2(sin²α - sin²αcosα - cos²αsinα - sin²αcosα + cosαsinα)/cosα

Simplifying:

Cds4α + 1 = 2sin²α - 4sin²αcosα - 2sin²αcosα