化简三角函数表达式
? TrigFactor[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1)]|Simplify[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1),Trig→True]|TrigReduce[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1)]|Cancel[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1),Trig→True]
? TrigFactor[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1)]|Simplify[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1),Trig→True]|TrigReduce[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1)]|Cancel[(Sin[2x]-Cos[2x]+1)/(Sin[2x]+Cos[2x]+1),Trig→True]
举一反三
- 已知\( y = {x^3}\cos 2x \)，则\( y'' \)为（ ）. A: 0 B: \( 6x\cos 2x{\rm{ + }}12{x^2}\sin 2x - 4{x^3}\cos 2x \) C: \( 6x\cos 2x - 12{x^2}\sin 2x{\rm{ + }}4{x^3}\cos 2x \) D: \( 6x\cos 2x - 12{x^2}\sin 2x - 4{x^3}\cos 2x \)
- 设 $f(\sin x)=\cos2x+1$，则 $f(\cos x)=$（ ）. A: $\cos^2x$ B: $-2\cos^2x$ C: $-2\sin^2x$ D: $2-2\cos^2x$
- \( 2{\sin ^2}x + \cos 2x = \) ________. ______
- 函数\(y = {\sin ^2}x\)的导数为（ ）. A: \(\cos 2x\) B: \(\sin 2x\) C: \( - \sin 2x\) D: \(- \cos 2x\)
- 设$f(x)=x\sin (2x)$，则${{f}^{(5)}}(x)=$( )。 A: $16[2x\cos (2x)+5\sin (2x)]$ B: $16[5\sin (2x)-2x\cos (2x)]$ C: $x\cos (2x)+5\sin (2x)$ D: $5\sin (2x)-x\cos (2x)$