Question

Sin2a=(sina+cosa)2-1

Answer 1

Ответ:-π/4

Пошаговое объяснение:

т.к. sin2α=2*sinα*cosα

2*sinα*cosα=(sinα+cosα)*2-1

2*sinα*cosα+1 =(sinα+cosα)*2

1=sin²α+cos²α  -  основное тригон. тождество

2*sinα*cosα+ sin²α+cos²α =(sinα+cosα)*2

собираем левую часть по формуле квадрат суммы

(sinα+cosα)²=(sinα+cosα)*2

(sinα+cosα)²- (sinα+cosα)*2=0. вынесем скобки за скобки

(sinα+cosα)*(sinα+cosα- 2)=0

(sinα+cosα)=0 или                          sinα+cosα- 2=0

tgα+1=0                                            sinα+cosα= 2 -быть не может

tgα=-1                                               ∅

α=-arctg1=-π/4

Answer 2

The equation you've written, Sin(2a) = (sin(a) + cos(a))^2 - 1, is not accurate. The correct identity for Sin(2a) is:

Sin(2a) = 2 * sin(a) * cos(a)

This is known as the double angle identity for sine. It expresses the sine of twice an angle (2a) in terms of the sines and cosines of the angle a.

If you expand the right side of your equation, you'll get a different result:

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

Using the Pythagorean identity (sin^2(a) + cos^2(a) = 1), this simplifies to:

2 * sin(a) * cos(a)

So, the correct identity is Sin(2a) = 2 * sin(a) * cos(a), not the equation you provided.