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受欢迎的三角函数 >

证明 sin(t)tan(t)=(1-cos^2(t))/(cos(t))

解答

证明

sin (t) tan (t) = \frac{1 - cos ^{2} ( t )}{cos ( t )}

解答

真

求解步骤

sin (t) tan (t) = \frac{1 - cos ^{2} ( t )}{cos ( t )}

调整左侧

sin (t) tan (t)

用 sin, cos 表示

sin (t) tan (t)

使用基本三角恒等式:

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

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

化简

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

sin (t) \frac{sin ( t )}{cos ( t )}

分式相乘:

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

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

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

sin (t) sin (t)

使用指数法则:

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

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

= sin^{1 + 1} (t)

数字相加:

1 + 1 = 2

= sin^{2} (t)

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

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

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

使用三角恒等式改写

\frac{sin ^{2} ( t )}{cos ( t )}

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} ( t )}{cos ( t )}

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

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 (csc(x))/(sin(x))-(cot(x))/(tan(x))=1

p ro v e \frac{csc ( x )}{sin ( x )} - \frac{cot ( x )}{tan ( x )} = 1

证明 (csc(x)-cot(x))/(sec(x)-1)=cot(x)

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

证明 (1+cos(x))/(1-cos(x))-(1-cos(x))/(1+cos(x))=4cot(x)csc(x)

p ro v e \frac{1 + cos ( x )}{1 - cos ( x )} - \frac{1 - cos ( x )}{1 + cos ( x )} = 4 cot (x) csc (x)

证明 cos(z+w)=cos(z)cos(w)-sin(z)sin(w)

p ro v e cos (z + w) = cos (z) cos (w) - sin (z) sin (w)

证明 tan(pi-θ)=-tan(θ)

p ro v e tan (π - θ) = - tan (θ)