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受欢迎的三角函数 >

证明 csc(u)-sin(u)=cot(u)cos(u)

解答

证明

csc (u) - sin (u) = cot (u) cos (u)

解答

真

求解步骤

csc (u) - sin (u) = cot (u) cos (u)

调整左侧

csc (u) - sin (u)

用 sin, cos 表示

csc (u) - sin (u)

使用基本三角恒等式:

csc (x) = \frac{1}{sin ( x )}

= \frac{1}{sin ( u )} - sin (u)

化简

\frac{1}{sin ( u )} - sin (u) : \frac{1 - sin ^{2} ( u )}{sin ( u )}

\frac{1}{sin ( u )} - sin (u)

将项转换为分式:

sin (u) = \frac{sin ( u ) sin ( u )}{sin ( u )}

= \frac{1}{sin ( u )} - \frac{sin ( u ) sin ( u )}{sin ( u )}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 - sin ( u ) sin ( u )}{sin ( u )}

1 - sin (u) sin (u) = 1 - sin^{2} (u)

1 - sin (u) sin (u)

sin (u) sin (u) = sin^{2} (u)

sin (u) sin (u)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

sin (u) sin (u) = sin^{1 + 1} (u)

= sin^{1 + 1} (u)

数字相加:

1 + 1 = 2

= sin^{2} (u)

= 1 - sin^{2} (u)

= \frac{1 - sin ^{2} ( u )}{sin ( u )}

= \frac{1 - sin ^{2} ( u )}{sin ( u )}

= \frac{1 - sin ^{2} ( u )}{sin ( u )}

使用三角恒等式改写

\frac{1 - sin ^{2} ( u )}{sin ( u )}

使用毕达哥拉斯恒等式:

1 = cos^{2} (x) + sin^{2} (x)

1 - sin^{2} (x) = cos^{2} (x)

= \frac{cos ^{2} ( u )}{sin ( u )}

= \frac{cos ^{2} ( u )}{sin ( u )}

使用三角恒等式改写

= cos (u) \frac{cos ( u )}{sin ( u )}

使用基本三角恒等式:

\frac{cos ( x )}{sin ( x )} = cot (x)

cos (u) cot (u)

cos (u) cot (u)

= cot (u) cos (u)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

流行的例子

证明 (tan(x))/(cot(x))=tan^2(x)

p ro v e \frac{tan ( x )}{cot ( x )} = tan^{2} (x)

证明-1/2 =sin(-pi/(12))sqrt(2+\sqrt{3)}

p ro v e - \frac{1}{2} = sin (- \frac{π}{12}) 2 + 3

证明 sin(4x)=2sin(x)(cos(x)+cos(3x))

p ro v e sin (4 x) = 2 sin (x) (cos (x) + cos (3 x))

证明 1-cos^2(x)-cos^2(x)=-cos(2x)

p ro v e 1 - cos^{2} (x) - cos^{2} (x) = - cos (2 x)

证明 sin^2(x)=(sec(x)sin(x))/(tan(x)+cot(x))

p ro v e sin^{2} (x) = \frac{sec ( x ) sin ( x )}{tan ( x ) + cot ( x )}