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受欢迎的三角函数 >

证明 csc(y)cos^2(y)+sin(y)=csc(y)

解答

证明

csc (y) cos^{2} (y) + sin (y) = csc (y)

解答

真

求解步骤

csc (y) cos^{2} (y) + sin (y) = csc (y)

调整左侧

csc (y) cos^{2} (y) + sin (y)

用 sin, cos 表示

sin (y) + cos^{2} (y) csc (y)

使用基本三角恒等式:

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

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

化简

sin (y) + cos^{2} (y) \frac{1}{sin ( y )} : \frac{sin ^{2} ( y ) + cos ^{2} ( y )}{sin ( y )}

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

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

cos^{2} (y) \frac{1}{sin ( y )}

分式相乘:

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

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

乘以:

1 \cdot cos^{2} (y) = cos^{2} (y)

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

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

将项转换为分式:

sin (y) = \frac{sin ( y ) sin ( y )}{sin ( y )}

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

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{sin ( y ) sin ( y ) + cos ^{2} ( y )}{sin ( y )}

sin (y) sin (y) + cos^{2} (y) = sin^{2} (y) + cos^{2} (y)

sin (y) sin (y) + cos^{2} (y)

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

sin (y) sin (y)

使用指数法则:

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

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

= sin^{1 + 1} (y)

数字相加:

1 + 1 = 2

= sin^{2} (y)

= sin^{2} (y) + cos^{2} (y)

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

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

= \frac{1}{sin ( y )}

= \frac{1}{sin ( y )}

使用三角恒等式改写

使用基本三角恒等式:

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

\frac{1}{\frac{1}{c s c ( y )}}

化简

\frac{1}{\frac{1}{c s c ( y )}}

使用分式法则:

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

= \frac{csc ( y )}{1}

使用法则

\frac{a}{1} = a

= csc (y)

csc (y)

csc (y)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

流行的例子

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p ro v e sin (x) - cos (x) = 1 - sin (2 x)

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p ro v e sin (3 A) = 3 sin (A) - 4 sin^{3} (A)