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受欢迎的三角函数 >

证明 ((cos^2(x)))/((1-sin(x)))=1+sin(x)

解答

证明

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

解答

真

求解步骤

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

调整左侧

\frac{cos ^{2} ( x )}{1 - sin ( x )}

使用三角恒等式改写

\frac{cos ^{2} ( x )}{1 - sin ( x )}

使用毕达哥拉斯恒等式:

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

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

= \frac{1 - sin ^{2} ( x )}{1 - sin ( x )}

化简

\frac{1 - sin ^{2} ( x )}{1 - sin ( x )} : sin (x) + 1

\frac{1 - sin ^{2} ( x )}{1 - sin ( x )}

分解

1 - sin^{2} (x) : - (sin (x) + 1) (sin (x) - 1)

1 - sin^{2} (x)

因式分解出通项

- 1

= - (sin^{2} (x) - 1)

分解

sin^{2} (x) - 1 : (sin (x) + 1) (sin (x) - 1)

sin^{2} (x) - 1

将

1

改写为

1^{2}

= sin^{2} (x) - 1^{2}

使用平方差公式:

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

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

= (sin (x) + 1) (sin (x) - 1)

= - (sin (x) + 1) (sin (x) - 1)

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

消掉

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

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

- sin (x) + 1 = - (sin (x) - 1)

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

整理后得

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

约分:

sin (x) - 1

= sin (x) + 1

= sin (x) + 1

= sin (x) + 1

= sin (x) + 1

= 1 + sin (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 8csc^2(x)-3cot^2(x)=3+5csc^2(x)

p ro v e 8 csc^{2} (x) - 3 cot^{2} (x) = 3 + 5 csc^{2} (x)

证明 sin^2(t)=(sin(t))^2

p ro v e sin^{2} (t) = (sin (t))^{2}

sin(2x)-2cos(x)=0,(0,2pi)

sin (2 x) - 2 cos (x) = 0, (0, 2 π)

sin (\frac{( 5 π )}{3})

sin (\frac{( 5 π )}{2})