Question

Sina*cos2a-cos(-a)*sin(-2a​

Answer 1

$\sin( \alpha ) \times \cos(2 \alpha ) - \cos( - \alpha ) \times \sin( - 2 \alpha ) \\ \\ \sin( \alpha ) \times \cos(2 \alpha ) + \cos( \alpha ) \times \sin(2 \alpha ) \\ \\ \sin(3 \alpha )$

$otvet( \sin(3 \alpha )$

~•~•~•ZLOY_TIGROVSKIY~•~•~•

Answer 2

To simplify the expression Sina*cos2a-cos(-a)*sin(-2a), we need to apply trigonometric identities. Let's start step by step:

1. Double-Angle Identity: cos(2a) = 2*cos^2(a) - 1

2. Negative Angle Identity: cos(-a) = cos(a)

3. Negative Angle Identity: sin(-a) = -sin(a)

4. Double-Angle Identity: sin(-2a) = -2*sin(a)*cos(a)

Now, we can rewrite the expression using these identities:

Sina * cos2a - cos(-a) * sin(-2a)

= Sina * (2cos^2(a) - 1) - cos(a) * (-2sin(a)*cos(a))

Now, distribute the terms:

= 2Sinacos^2(a) - Sina - (-2*cos(a)*sin(a)*cos(a))

= 2Sinacos^2(a) - Sina + 2*cos(a)*sin(a)*cos(a)

Now, let's factor out common terms:

= Sina * (2cos^2(a) - 1) + 2cos(a)*sin(a)*cos(a)

Now, we can use the Pythagorean Identity: 1 - sin^2(a) = cos^2(a)

= Sina * (2*(1 - sin^2(a)) - 1) + 2*cos(a)*sin(a)*cos(a)

Simplify further:

= Sina * (2 - 2sin^2(a) - 1) + 2cos(a)*sin(a)*cos(a)

= Sina * (1 - 2sin^2(a)) + 2cos(a)*sin(a)*cos(a)

Now, we can use the double-angle identity for sine: sin(2a) = 2*sin(a)*cos(a)

= Sina * (1 - 2*sin^2(a)) + sin(2a)*cos(a)

And that's the simplified expression.