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受欢迎的三角函数 >

证明 1+1/(tan^2(θ))= 1/(sin^2(θ))

解答

证明

1 + \frac{1}{tan ^{2} ( θ )} = \frac{1}{sin ^{2} ( θ )}

解答

真

求解步骤

1 + \frac{1}{tan ^{2} ( θ )} = \frac{1}{sin ^{2} ( θ )}

调整左侧

1 + \frac{1}{tan ^{2} ( θ )}

用 sin, cos 表示

1 + \frac{1}{tan ^{2} ( θ )}

使用基本三角恒等式:

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

= 1 + \frac{1}{( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}

化简

1 + \frac{1}{( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}} : \frac{sin ^{2} ( θ ) + cos ^{2} ( θ )}{sin ^{2} ( θ )}

1 + \frac{1}{( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}

\frac{1}{( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}} = \frac{cos ^{2} ( θ )}{sin ^{2} ( θ )}

\frac{1}{( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}

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

(\frac{sin ( θ )}{cos ( θ )})^{2}

使用指数法则:

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

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

= \frac{1}{\frac{s i n ^{2} ( θ )}{c o s ^{2} ( θ )}}

使用分式法则:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

= 1 + \frac{cos ^{2} ( θ )}{sin ^{2} ( θ )}

将项转换为分式:

1 = \frac{1 sin ^{2} ( θ )}{sin ^{2} ( θ )}

= \frac{1 \cdot sin ^{2} ( θ )}{sin ^{2} ( θ )} + \frac{cos ^{2} ( θ )}{sin ^{2} ( 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{1 \cdot sin ^{2} ( θ ) + cos ^{2} ( θ )}{sin ^{2} ( θ )}

乘以:

1 \cdot sin^{2} (θ) = sin^{2} (θ)

= \frac{sin ^{2} ( θ ) + cos ^{2} ( θ )}{sin ^{2} ( θ )}

= \frac{sin ^{2} ( θ ) + cos ^{2} ( θ )}{sin ^{2} ( θ )}

= \frac{cos ^{2} ( θ ) + sin ^{2} ( θ )}{sin ^{2} ( θ )}

使用三角恒等式改写

\frac{cos ^{2} ( θ ) + sin ^{2} ( θ )}{sin ^{2} ( θ )}

使用毕达哥拉斯恒等式:

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

= \frac{1}{sin ^{2} ( θ )}

= \frac{1}{sin ^{2} ( θ )}

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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