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受欢迎的三角函数 >

证明 cot(θ)(sin(θ)+tan(θ))=cos(θ)+1

解答

证明

cot (θ) (sin (θ) + tan (θ)) = cos (θ) + 1

解答

真

求解步骤

cot (θ) (sin (θ) + tan (θ)) = cos (θ) + 1

调整左侧

cot (θ) (sin (θ) + tan (θ))

用 sin, cos 表示

(sin (θ) + tan (θ)) cot (θ)

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

化简

(sin (θ) + \frac{sin ( θ )}{cos ( θ )}) \frac{cos ( θ )}{sin ( θ )} : cos (θ) + 1

(sin (θ) + \frac{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 ) ( sin ( θ ) + \frac{s i n ( θ )}{c o s ( θ )} )}{sin ( θ )}

化简

sin (θ) + \frac{sin ( θ )}{cos ( θ )} : \frac{sin ( θ ) cos ( θ ) + sin ( θ )}{cos ( θ )}

sin (θ) + \frac{sin ( θ )}{cos ( θ )}

将项转换为分式:

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

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

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{sin ( θ ) cos ( θ ) + sin ( θ )}{cos ( θ )}

= \frac{\frac{s i n ( θ ) c o s ( θ ) + s i n ( θ )}{c o s ( θ )} cos ( θ )}{sin ( θ )}

乘

cos (θ) \frac{sin ( θ ) cos ( θ ) + sin ( θ )}{cos ( θ )} : sin (θ) cos (θ) + sin (θ)

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

分式相乘:

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

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

约分:

cos (θ)

= sin (θ) cos (θ) + sin (θ)

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

因式分解出通项

sin (θ)

= \frac{sin ( θ ) ( cos ( θ ) + 1 )}{sin ( θ )}

约分:

sin (θ)

= cos (θ) + 1

= cos (θ) + 1

= 1 + cos (θ)

= cos (θ) + 1

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 (2-sin^2(x))csc^2(x)=cot^2(x)

p ro v e (2 - sin^{2} (x)) csc^{2} (x) = cot^{2} (x)

证明 1/(tan(A))+tan(A)= 2/(sin(2A))

p ro v e \frac{1}{tan ( A )} + tan (A) = \frac{2}{sin ( 2 A )}

证明 1+sin(θ)=cos(θ)

p ro v e 1 + sin (θ) = cos (θ)

证明 1+((tan^2(x)))/(1+sec(x))=sec(x)

p ro v e 1 + \frac{( tan ^{2} ( x ) )}{1 + sec ( x )} = sec (x)

证明 csc^2(x)*cos^2(x)=cot^2(x)

p ro v e csc^{2} (x) \cdot cos^{2} (x) = cot^{2} (x)