证明 (sin(3x)-sin(5x))/(2cos(4x))=-sin(x)

解答

证明

\frac{sin ( 3 x ) - sin ( 5 x )}{2 cos ( 4 x )} = - sin (x)

真

求解步骤

\frac{sin ( 3 x ) - sin ( 5 x )}{2 cos ( 4 x )} = - sin (x)

调整左侧

\frac{sin ( 3 x ) - sin ( 5 x )}{2 cos ( 4 x )}

使用三角恒等式改写

\frac{sin ( 3 x ) - sin ( 5 x )}{2 cos ( 4 x )}

使用和差化积恒等式:

sin (s) - sin (t) = 2 sin (\frac{s - t}{2}) cos (\frac{s + t}{2})

= \frac{2 sin ( \frac{3 x - 5 x}{2} ) cos ( \frac{3 x + 5 x}{2} )}{2 cos ( 4 x )}

化简

\frac{2 sin ( \frac{3 x - 5 x}{2} ) cos ( \frac{3 x + 5 x}{2} )}{2 cos ( 4 x )} : - sin (x)

\frac{2 sin ( \frac{3 x - 5 x}{2} ) cos ( \frac{3 x + 5 x}{2} )}{2 cos ( 4 x )}

2 sin (\frac{3 x - 5 x}{2}) cos (\frac{3 x + 5 x}{2}) = 2 sin (- \frac{2 x}{2}) cos (\frac{8 x}{2})

2 sin (\frac{3 x - 5 x}{2}) cos (\frac{3 x + 5 x}{2})

\frac{3 x - 5 x}{2} = - \frac{2 x}{2}

\frac{3 x - 5 x}{2}

p ro v e tan (a) + cot (a) = 2 csc (2 a)

p ro v e \frac{sec ( t ) - 1}{tan ( t )} = \frac{tan ( t )}{sec ( t ) + 1}

p ro v e 3 cos^{2} (x) - 3 sin^{2} (x) = 3 - 6 sin^{2} (x)

p ro v e sec^{4} (x) = \frac{1}{cos ^{4} ( x )}

p ro v e \frac{1}{tan ( x )} = cot (x)