Well, maths is about generalising problems, so let's solve instead [math](\sin x +\cos x)^2[/math], and we'll plug in the [math]18^{\circ}[/math] later.
So, we will first expand the binomial:
[math](\sin x + \cos x)^2 = \sin^2x + 2\sin x\cos x + \cos^2x[/math]
Since [math]\sin^2x + \cos^2x = 1[/math] we can write the expression a...
Không có nhận xét nào:
Đăng nhận xét