Question

(Sin x + Cos x)^{2}=1+Sin x Cos x

Answer 1

Ответ:

Объяснение:

sin^2x + 2sinx*cosx + cos^2x = 1 + sinx*cosx

1 + 2sinx*cosx = 1 + sinx*cosx

2sinx*cosx - sinx*cosx = 0  

sinx*cosx = 0  

1) sinx = 0  

x = pik, k ∈ Z

2) cosx = 0  

x = pi/2+ pik, k∈Z

Answer 2

To prove the trigonometric identity $(\sin x + \cos x)^2 = 1 + \sin x \cos x$, you can follow these steps:

Start with the left-hand side (LHS) of the identity: $(\sin x + \cos x)^2$

Use the formula for squaring a binomial: $(\sin x + \cos x)(\sin x + \cos x)$

Now, apply the distributive property to expand the expression: $\sin^2 x + \sin x \cos x + \cos x \sin x + \cos^2 x$

Notice that $\sin x \cos x$ and $\cos x \sin x$ are the same, so we can simplify the expression: $\sin^2 x + 2\sin x \cos x + \cos^2 x$

Now, recall the Pythagorean identity $\sin^2 x + \cos^2 x = 1$. You can replace $\sin^2 x + \cos^2 x$ with 1: $1 + 2\sin x \cos x$

Finally, simplify further: $1 + 2\sin x \cos x$

This is the right-hand side (RHS) of the identity. Therefore, we have shown that $(\sin x + \cos x)^2 = 1 + \sin x \cos x$, which is the desired trigonometric identity.