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受欢迎的三角函数 >

证明 cos(x)tan^3(x)=sin(x)tan^2(x)

解答

证明

cos (x) tan^{3} (x) = sin (x) tan^{2} (x)

解答

真

求解步骤

cos (x) tan^{3} (x) = sin (x) tan^{2} (x)

调整左侧

cos (x) tan^{3} (x)

用 sin, cos 表示

cos (x) tan^{3} (x)

使用基本三角恒等式:

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

= cos (x) (\frac{sin ( x )}{cos ( x )})^{3}

化简

cos (x) (\frac{sin ( x )}{cos ( x )})^{3} : \frac{sin ^{3} ( x )}{cos ^{2} ( x )}

cos (x) (\frac{sin ( x )}{cos ( x )})^{3}

(\frac{sin ( x )}{cos ( x )})^{3} = \frac{sin ^{3} ( x )}{cos ^{3} ( x )}

(\frac{sin ( x )}{cos ( x )})^{3}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{sin ^{3} ( x )}{cos ^{3} ( x )}

= \frac{sin ^{3} ( x )}{cos ^{3} ( x )} cos (x)

分式相乘:

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

= \frac{sin ^{3} ( x ) cos ( x )}{cos ^{3} ( x )}

约分:

cos (x)

= \frac{sin ^{3} ( x )}{cos ^{2} ( x )}

= \frac{sin ^{3} ( x )}{cos ^{2} ( x )}

= \frac{sin ^{3} ( x )}{cos ^{2} ( x )}

使用三角恒等式改写

= \frac{sin ^{2} ( x )}{cos ( x )} \cdot \frac{sin ( x )}{cos ( x )}

使用基本三角恒等式:

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

\frac{sin ^{2} ( x ) tan ( x )}{cos ( x )}

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

使用基本三角恒等式:

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

sin (x) tan (x) tan (x)

化简

sin (x) tan (x) tan (x) : tan^{2} (x) sin (x)

sin (x) tan (x) tan (x)

使用指数法则:

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

tan (x) tan (x) = tan^{1 + 1} (x)

= sin (x) tan^{1 + 1} (x)

数字相加:

1 + 1 = 2

= sin (x) tan^{2} (x)

tan^{2} (x) sin (x)

tan^{2} (x) sin (x)

= sin (x) tan^{2} (x)

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 sin^2(0)=4(cos(-0)-1)

p ro v e sin^{2} (0) = 4 (cos (- 0) - 1)

证明 sec(2a)+tan(2a)=(cos(a)+sin(a))/(cos(a)-sin(a))

p ro v e sec (2 a) + tan (2 a) = \frac{cos ( a ) + sin ( a )}{cos ( a ) - sin ( a )}

证明 sec^2(x)= 1/(1-sin^2(x))

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

证明 cos(-x)+sin(-x)=cos(x)-sin(x)

p ro v e cos (- x) + sin (- x) = cos (x) - sin (x)

证明 (1-cos(2θ))/(cos(2θ)+1)=tan^2(θ)

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