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受欢迎的三角函数 >

证明 cos(θ)cot(θ)= 1/(sin(θ))-sin(θ)

解答

证明

cos (θ) cot (θ) = \frac{1}{sin ( θ )} - sin (θ)

解答

真

求解步骤

cos (θ) cot (θ) = \frac{1}{sin ( θ )} - sin (θ)

调整左侧

cos (θ) cot (θ)

用 sin, cos 表示

cos (θ) cot (θ)

使用基本三角恒等式:

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

= cos (θ) \frac{cos ( θ )}{sin ( θ )}

化简

cos (θ) \frac{cos ( θ )}{sin ( θ )} : \frac{cos ^{2} ( θ )}{sin ( θ )}

cos (θ) \frac{cos ( θ )}{sin ( θ )}

分式相乘:

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

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

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

使用指数法则:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

数字相加:

1 + 1 = 2

= cos^{2} (θ)

= \frac{cos ^{2} ( θ )}{sin ( θ )}

= \frac{cos ^{2} ( θ )}{sin ( θ )}

= \frac{cos ^{2} ( θ )}{sin ( θ )}

使用三角恒等式改写

\frac{cos ^{2} ( θ )}{sin ( θ )}

使用毕达哥拉斯恒等式:

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

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

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

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

乘开

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

\frac{1 - sin ^{2} ( θ )}{sin ( θ )}

使用分式法则:

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

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

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

消掉

\frac{sin ^{2} ( θ )}{sin ( θ )} : sin (θ)

\frac{sin ^{2} ( θ )}{sin ( θ )}

约分:

sin (θ)

= sin (θ)

= \frac{1}{sin ( θ )} - sin (θ)

= \frac{1}{sin ( θ )} - sin (θ)

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 1-cos(θ)sin(θ)tan(θ)=cos^2(θ)

p ro v e 1 - cos (θ) sin (θ) tan (θ) = cos^{2} (θ)

证明 (1-sin^2(A))/(1-cos^2(A))=cot^2(A)

p ro v e \frac{1 - sin ^{2} ( A )}{1 - cos ^{2} ( A )} = cot^{2} (A)

证明 (sin(60))/(cos(60))=tan(60)

p ro v e \frac{sin ( 6 0 ^{\circ} )}{cos ( 6 0 ^{\circ} )} = tan (6 0^{\circ})

证明 (tan(x)*cot(x))/(csc(x))=sin(x)

p ro v e \frac{tan ( x ) \cdot cot ( x )}{csc ( x )} = sin (x)

证明 1/(1+cot^2(a))=sin^2(a)

p ro v e \frac{1}{1 + cot ^{2} ( a )} = sin^{2} (a)