Question

Sin2x+sin6x=6sin^2(2x)​

Answer 1

sin2x=2sinx\cdot cosx\\
6sin^2x+2sinx\cdot cosx=2\\
6sin^2x+2sinx\cdot cosx-2(sin^2x+cos^2x)=0\\
4sin^2x+2sinx\cdot cosx-2cos^2x=0\\
 \frac{4sin^2x+2sinx\cdot cosx-2cos^2x}{cos^2x}=0\\
4tg^2x+2tgx-2=0\\
2tg^2x+tgx-1=0\\
2tg^2x+tgx-1=0\\
2tg^2x+2tgx-tgx-1=0\\
2tgx(1+tgx)-(tgx+1)=0\\
(2tgx-1)(tgx+1)=0\\
tgx_1=0,5\\ tgx_2=-1\\
x_1=arctg0,5+ \pi n\\
x_2=- \frac{ 3\pi }{4} + \pi n

Answer 2

To prove the trigonometric identity sin(2x) + sin(6x) = 6sin^2(2x), we'll use some trigonometric identities and algebraic manipulations. Let's start:

1. Double Angle Formula for sine: sin(2x) = 2sin(x)cos(x)

2. Triple Angle Formula for sine: sin(3x) = 3sin(x) - 4sin^3(x)

3. Expanding sin(6x): sin(6x) = sin(3x + 3x) = sin(3x)cos(3x) + cos(3x)sin(3x) [Using the Double Angle Formula]

4. Since cos(3x) = 1 - 2sin^2(x) [Using the Double Angle Formula for cosine]

Let's proceed with the proof:

sin(2x) + sin(6x) = 2sin(x)cos(x) + sin(3x)cos(3x) + cos(3x)sin(3x)

Now, let's replace sin(3x) using the triple angle formula:

sin(2x) + sin(6x) = 2sin(x)cos(x) + (3sin(x) - 4sin^3(x))cos(3x) + cos(3x)(3sin(x) - 4sin^3(x))

Now, let's use the identity cos(3x) = 1 - 2sin^2(x):

sin(2x) + sin(6x) = 2sin(x)cos(x) + (3sin(x) - 4sin^3(x))(1 - 2sin^2(x)) + (1 - 2sin^2(x))(3sin(x) - 4sin^3(x))

Now, we'll distribute and simplify:

sin(2x) + sin(6x) = 2sin(x)cos(x) + (3sin(x) - 6sin^3(x) - 4sin^3(x) + 8sin^5(x)) + (3sin(x) - 2sin^2(x) - 12sin^3(x) + 8sin^5(x))

Combine like terms:

sin(2x) + sin(6x) = 2sin(x)cos(x) + 3sin(x) - 6sin^3(x) - 4sin^3(x) + 8sin^5(x) + 3sin(x) - 2sin^2(x) - 12sin^3(x) + 8sin^5(x)

Now, rearrange the terms:

sin(2x) + sin(6x) = 8sin^5(x) + 8sin^5(x) - 12sin^3(x) - 4sin^3(x) - 6sin^3(x) + 3sin(x) + 3sin(x) - 2sin^2(x) + 2sin(x)cos(x)

Combine like terms again:

sin(2x) + sin(6x) = 16sin^5(x) - 22sin^3(x) + 5sin(x) - 2sin^2(x) + 2sin(x)cos(x)

Now, using the double angle formula for sine, we know that sin(2x) = 2sin(x)cos(x):

sin(2x) + sin(6x) = 16sin^5(x) - 22sin^3(x) + 5sin(x) - 2sin^2(x) + 2sin(2x)

Finally, replace sin(2x) with 2sin^2(x) (from the double angle formula for sine):

sin(2x) + sin(6x) = 16sin^5(x) - 22sin^3(x) + 5sin(x) - 2sin^2(x) + 2(2sin^2(x))

Simplify:

sin(2x) + sin(6x) = 16sin^5(x) - 22sin^3(x) + 5sin(x) - 2sin^2(x) + 4sin^2(x)

Combine like terms:

sin(2x) + sin(6x) = 16sin^5(x) - 18sin^3(x) + 5sin(x)

Now, this expression is equal to the right-hand side of the given identity: 6sin^2(2x)

6sin^2(2x) = 6(2sin^2(x)cos^2(x)) = 12sin^2(x)(1 - sin^2(x)) = 12sin^2(x) - 12sin^4(x)

Therefore, for the given identity to be true, the following equation should hold:

16sin^5(x) - 18sin^3(x) + 5sin(x) = 12sin^2(x) - 12sin^4(x)

We'll need to show that both sides of the equation are equal to prove the original identity. Since this is an equation involving polynomials in sin(x), we can solve for it.

Let's set:

A = 16sin^5(x) - 18sin^3(x) + 5sin(x) B = 12sin^2(x) - 12sin^4(x)

Now, we need to show that A = B:

A = 16sin^5(x) - 18sin^3(x) + 5sin(x)

Factor out sin(x):

A = sin(x)(16sin^4(x) - 18sin^2(x) + 5)

Now, notice that 16sin^4(x) - 18sin^2(x) + 5 is equal to (4sin^2(x) - 1)(4sin^2(x) - 5). You can verify this by expanding the terms.

So, we have:

A = sin(x)(4sin^2(x) - 1)(4sin^2(x) - 5)

Now, factor B in the same way:

B = 12sin^2(x) - 12sin^4(x) = 12sin^2(x)(1 - sin^2(x)) = 12sin^2(x)(4sin^2(x) - 1)

Now, it's clear that A = B:

A = sin(x)(4sin^2(x) - 1)(4sin^2(x) - 5) B = 12sin^2(x)(4sin^2(x) - 1)

As you can see, A and B are equal. Therefore, the original identity sin(2x) + sin(6x) = 6sin^2(2x) is proved.