Question

Sin2x*cosx+sinx*cos2x=корень из 2/2

Answer 1

$\displaystyle sin2xcosx+sinxcos2x=\frac{\sqrt2}{2}\\sin(2x+x)=\frac{\sqrt2}{2}\\sin(3x)=\frac{\sqrt2}{2}\\3x_1=\frac{\pi}{4}+2\pi n;3x_2=\frac{3\pi}{4}+2\pi n;n\in Z\\x_1=\frac{\pi}{12}+\frac{2\pi n}{3};x_2=\frac{3\pi}{12}+\frac{2\pi n}{3};n\in Z$

Answer 2

To verify the given trigonometric expression, we'll use some trigonometric identities. First, let's recall the double angle identities:

1. Sin(2x) = 2 * Sin(x) * Cos(x)
2. Cos(2x) = Cos^2(x) - Sin^2(x) = 2 * Cos^2(x) - 1 = 1 - 2 * Sin^2(x)

Now, we'll simplify the expression step by step:

Sin(2x) * Cos(x) + Sin(x) * Cos(2x)

Using the double angle identity for Sin(2x), we get:

(2 * Sin(x) * Cos(x)) * Cos(x) + Sin(x) * Cos(2x)

Next, using the double angle identity for Cos(2x), we get:

(2 * Sin(x) * Cos(x)) * Cos(x) + Sin(x) * (1 - 2 * Sin^2(x))

Now, distribute the terms:

2 * Sin(x) * Cos^2(x) + Sin(x) - 2 * Sin^3(x)

Now, notice that 2 * Sin(x) * Cos^2(x) can be further simplified using the identity:

Cos^2(x) = 1 - Sin^2(x)

Substitute this into the expression:

2 * Sin(x) * (1 - Sin^2(x)) + Sin(x) - 2 * Sin^3(x)

Now, we have a common term of Sin(x) in both terms. Factor it out:

Sin(x) * (2 - 2 * Sin^2(x) + 1 - 2 * Sin^2(x))

Now, combine like terms:

Sin(x) * (3 - 4 * Sin^2(x))

To find the value of the expression, we need to solve for Sin(x). Set the expression equal to the given value of √2/2:

Sin(x) * (3 - 4 * Sin^2(x)) = √2/2

Now, we know that Sin(45°) = √2/2. Therefore, set the expression inside the parentheses equal to 1 (since Sin(45°) = 1):

3 - 4 * Sin^2(x) = 1

Now, solve for Sin^2(x):

4 * Sin^2(x) = 2

Sin^2(x) = 1/2

Taking the square root of both sides:

Sin(x) = ±√(1/2) = ±√2/2

So, Sin(x) can be either √2/2 or -√2/2.

Therefore, the original expression Sin(2x) * Cos(x) + Sin(x) * Cos(2x) is equal to √2/2 for certain values of x (when Sin(x) = √2/2) but may not be true for all values of x.