Question

(cos2x-sin4x)/sin2x=ctg2x

Answer 1

$\frac{cos2x-sin4x}{sin2x} =ctg2x$
$1. \frac{cos2x-sin4x}{sin2x}= \frac{cos2x}{sin2x}- \frac{sin4x}{sin2x}=ctg2x- \frac{2sin2x*cos2x}{sin2x}=ctg2x-2cos2x$
$2. ctg2x-2cos2x=ctg2x 2cos2x=0. cos2x=0$
$2x= \frac{ \pi }{2} + \pi n, n$∈Z
$x= \frac{ \pi }{4} + \frac{ \pi n}{2} , n$∈Z

Answer 2

To prove the identity (cos2x - sin4x)/sin2x = cot(2x), we can manipulate the left-hand side of the equation using trigonometric identities.

Let's start with the left-hand side:

(cos2x - sin4x)/sin2x

We'll focus on simplifying the numerator first.

1. Applying the double angle formula for cosine: cos(2x) = cos^2(x) - sin^2(x) The numerator becomes: (cos^2(x) - sin^2(x) - sin4x)

2. Recognizing that sin(4x) can be expressed in terms of double angles: sin(4x) = 2sin(2x)cos(2x)

Substituting sin(4x) in the numerator:

(cos^2(x) - sin^2(x) - 2sin(2x)cos(2x))

Now let's simplify the denominator:

sin2x = 2sin(x)cos(x)

Now we can rewrite the equation:

[(cos^2(x) - sin^2(x) - 2sin(2x)cos(2x))] / [2sin(x)cos(x)]

Next, we'll simplify the numerator further:

1. Rearranging the terms in the numerator to group like terms: (cos^2(x) - sin^2(x)) - 2sin(2x)cos(2x)

2. Recognizing the difference of squares identity: (cos^2(x) - sin^2(x)) = cos(2x)

The numerator becomes:

(cos(2x) - 2sin(2x)cos(2x))

Now, we can substitute back into the equation:

[(cos(2x) - 2sin(2x)cos(2x))] / [2sin(x)cos(x)]

Factoring out common terms:

[cos(2x)(1 - 2sin(2x))] / [2sin(x)cos(x)]

Now, we'll simplify the expression inside the brackets:

1. Using the double angle formula for sine: sin(2x) = 2sin(x)cos(x)

(1 - 2sin(2x)) = 1 - 2(2sin(x)cos(x)) = 1 - 4sin(x)cos(x)

Substituting this back into the equation:

[cos(2x)(1 - 4sin(x)cos(x))] / [2sin(x)cos(x)]

Now, we can simplify further:

1. Canceling out a factor of 2 in the numerator and denominator:

[cos(2x)(1 - 4sin(x)cos(x))] / [sin(x)cos(x)]

2. Recognizing that cos(2x)/sin(x) = cot(x):

[cos(2x)(1 - 4sin(x)cos(x))] / [sin(x)cos(x)] = cos(2x) * cot(x)

So, we have:

(cos(2x) - 2sin(2x)cos(2x)) / (2sin(x)cos(x)) = cos(2x) * cot(x)

Therefore, we've proven that (cos2x - sin4x)/sin2x = cot(2x).