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受欢迎的三角函数 >

证明 sec(α)-cos(α)=sin(α)tan(α)

解答

证明

sec (α) - cos (α) = sin (α) tan (α)

解答

真

求解步骤

sec (α) - cos (α) = sin (α) tan (α)

调整左侧

sec (α) - cos (α)

用 sin, cos 表示

- cos (α) + sec (α)

使用基本三角恒等式:

sec (x) = \frac{1}{cos ( x )}

= - cos (α) + \frac{1}{cos ( α )}

化简

- cos (α) + \frac{1}{cos ( α )} : \frac{- cos ^{2} ( α ) + 1}{cos ( α )}

- cos (α) + \frac{1}{cos ( α )}

将项转换为分式:

cos (α) = \frac{cos ( α ) cos ( α )}{cos ( α )}

= - \frac{cos ( α ) cos ( α )}{cos ( α )} + \frac{1}{cos ( α )}

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{- cos ( α ) cos ( α ) + 1}{cos ( α )}

- cos (α) cos (α) + 1 = - cos^{2} (α) + 1

- cos (α) cos (α) + 1

cos (α) cos (α) = cos^{2} (α)

cos (α) cos (α)

使用指数法则:

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

cos (α) cos (α) = cos^{1 + 1} (α)

= cos^{1 + 1} (α)

数字相加:

1 + 1 = 2

= cos^{2} (α)

= - cos^{2} (α) + 1

= \frac{- cos ^{2} ( α ) + 1}{cos ( α )}

= \frac{- cos ^{2} ( α ) + 1}{cos ( α )}

= \frac{1 - cos ^{2} ( α )}{cos ( α )}

调整右侧

sin (α) tan (α)

用 sin, cos 表示

sin (α) tan (α)

使用基本三角恒等式:

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

= sin (α) \frac{sin ( α )}{cos ( α )}

化简

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

sin (α) \frac{sin ( α )}{cos ( α )}

分式相乘:

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

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

sin (α) sin (α) = sin^{2} (α)

sin (α) sin (α)

使用指数法则:

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

sin (α) sin (α) = sin^{1 + 1} (α)

= sin^{1 + 1} (α)

数字相加:

1 + 1 = 2

= sin^{2} (α)

= \frac{sin ^{2} ( α )}{cos ( α )}

= \frac{sin ^{2} ( α )}{cos ( α )}

= \frac{sin ^{2} ( α )}{cos ( α )}

使用三角恒等式改写

\frac{sin ^{2} ( α )}{cos ( α )}

使用毕达哥拉斯恒等式:

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

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

= \frac{1 - cos ^{2} ( α )}{cos ( α )}

= \frac{1 - cos ^{2} ( α )}{cos ( α )}

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^2(θ)+6=sec^2(θ)+5

p ro v e tan^{2} (θ) + 6 = sec^{2} (θ) + 5

证明 sec(pi/2-x)=csc(x)

p ro v e sec (\frac{π}{2} - x) = csc (x)

证明 1/(csc(x)-1)+1/(csc(x)+1)=2tan(x)sec(x)

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

证明 (1+cos(t))/(1-cos(t))=(csc(t)+cot(t))^2

p ro v e \frac{1 + cos ( t )}{1 - cos ( t )} = (csc (t) + cot (t))^{2}

证明 sec^2(α)+csc^2(α)= 1/(sin^2(α)cos^2(α))

p ro v e sec^{2} (α) + csc^{2} (α) = \frac{1}{sin ^{2} ( α ) cos ^{2} ( α )}