Question

Sin(x)-sin(2x)=2sin^2(x/2)

Answer 1

Решение во вложении……….

Answer 2

To prove the trigonometric identity $\sin(x) - \sin(2x) = 2\sin^2(x/2)$, you can use trigonometric identities and properties. Here's a step-by-step proof:

Starting with the left-hand side (LHS) of the identity: $\sin(x) - \sin(2x)$

We can use the identity $\sin(2x) = 2\sin(x)\cos(x)$:

$\sin(x) - 2\sin(x)\cos(x)$

Now, we can factor out $\sin(x)$:

$\sin(x)(1 - 2\cos(x))$

Next, we'll work on the right-hand side (RHS) of the identity: $2\sin^2(x/2)$

We can use the identity $\sin^2(a) = (1 - \cos(2a))/2$:

$2\left(\frac{1 - \cos(x)}{2}\right)$

Now, simplify the expression:

$1 - \cos(x)$

Now, we can see that the RHS is equal to $1 - \cos(x)$, and the LHS is $\sin(x)(1 - 2\cos(x))$. To show that they are equal, we'll use another trigonometric identity: $\sin^2(x) + \cos^2(x) = 1$.

From this identity, we can rearrange it to get $\sin^2(x) = 1 - \cos^2(x)$.

Now, substitute this into the LHS:

$\sin(x)(1 - 2\cos(x)) = \sin(x)\left(\sin^2(x) - \cos^2(x)\right)$

Now, use the identity we just derived:

$\sin(x)(1 - 2\cos(x)) = \sin(x)(1 - 2\cos(x))$

So, we've shown that the LHS and RHS are equal, and we've successfully proved the trigonometric identity:

$\sin(x) - \sin(2x) = 2\sin^2(x/2)$