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受欢迎的三角函数 >

证明 (1-cos(b))(1+cos(b))= 1/(csc^2(b))

解答

证明

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

解答

真

求解步骤

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

调整右侧

\frac{1}{csc ^{2} ( b )}

用 sin, cos 表示

\frac{1}{csc ^{2} ( b )}

使用基本三角恒等式:

csc (x) = \frac{1}{sin ( x )}

= \frac{1}{( \frac{1}{s i n ( b )} ) ^{2}}

化简

\frac{1}{( \frac{1}{s i n ( b )} ) ^{2}} : sin^{2} (b)

\frac{1}{( \frac{1}{s i n ( b )} ) ^{2}}

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

(\frac{1}{sin ( b )})^{2}

使用指数法则:

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

= \frac{1}{\frac{1}{s i n ^{2} ( b )}}

使用分式法则:

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

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

使用法则

\frac{a}{1} = a

= sin^{2} (b)

= sin^{2} (b)

= sin^{2} (b)

使用三角恒等式改写

sin^{2} (b)

使用毕达哥拉斯恒等式:

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

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

= 1 - cos^{2} (b)

分解

1 - cos^{2} (b) : (1 + cos (b)) (1 - cos (b))

1 - cos^{2} (b)

使用平方差公式:

b^{2} - y^{2} = (b + y) (b - y)

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

= (1 + cos (b)) (1 - cos (b))

= (1 + cos (b)) (1 - cos (b))

= (1 + cos (b)) (1 - cos (b))

= (1 - cos (b)) (1 + cos (b))

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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