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受欢迎的三角函数 >

证明 (sin(x)+cos(x))/(cos(x))=1+tan(x)

解答

证明

\frac{sin ( x ) + cos ( x )}{cos ( x )} = 1 + tan (x)

解答

真

求解步骤

\frac{sin ( x ) + cos ( x )}{cos ( x )} = 1 + tan (x)

调整右侧

1 + tan (x)

用 sin, cos 表示

1 + tan (x)

使用基本三角恒等式:

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

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

化简

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

1 + \frac{sin ( x )}{cos ( x )}

将项转换为分式:

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

= \frac{1 \cdot cos ( x )}{cos ( x )} + \frac{sin ( x )}{cos ( x )}

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{1 \cdot cos ( x ) + sin ( x )}{cos ( x )}

乘以:

1 \cdot cos (x) = cos (x)

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

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

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

= \frac{sin ( x ) + cos ( x )}{cos ( 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 (1+sin(-x))/(cos(x)tan(x)-1)=-1

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

证明 sec(A)-(cos(A))/(1+sin(A))=tan(A)

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

证明 (1-tan^2(A))/(1+tan^2(A))=cos(2A)

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

证明 sin(pi/3)=cos(pi/6)

p ro v e sin (\frac{π}{3}) = cos (\frac{π}{6})

证明 cos^2(t)=((1+cos(2t)))/2

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