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受欢迎的三角函数 >

证明 (1-sin^4(a))/(cos^2(a))=2-cos^2(a)

解答

证明

\frac{1 - sin ^{4} ( a )}{cos ^{2} ( a )} = 2 - cos^{2} (a)

解答

真

求解步骤

\frac{1 - sin ^{4} ( a )}{cos ^{2} ( a )} = 2 - cos^{2} (a)

调整左侧

\frac{1 - sin ^{4} ( a )}{cos ^{2} ( a )}

使用三角恒等式改写

\frac{1 - sin ^{4} ( a )}{cos ^{2} ( a )}

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 ^{4} ( a )}{1 - sin ^{2} ( a )}

化简

\frac{1 - sin ^{4} ( a )}{1 - sin ^{2} ( a )} : sin^{2} (a) + 1

\frac{1 - sin ^{4} ( a )}{1 - sin ^{2} ( a )}

分解

1 - sin^{4} (a) : - (sin^{2} (a) + 1) (sin (a) + 1) (sin (a) - 1)

1 - sin^{4} (a)

因式分解出通项

- 1

= - (sin^{4} (a) - 1)

分解

sin^{4} (a) - 1 : (sin^{2} (a) + 1) (sin (a) + 1) (sin (a) - 1)

sin^{4} (a) - 1

将

sin^{4} (a) - 1

改写为

(sin^{2} (a))^{2} - 1^{2}

sin^{4} (a) - 1

将

1

改写为

1^{2}

= sin^{4} (a) - 1^{2}

使用指数法则:

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

sin^{4} (a) = (sin^{2} (a))^{2}

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

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

使用平方差公式:

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

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

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

分解

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

sin^{2} (a) - 1

将

1

改写为

1^{2}

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

使用平方差公式:

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

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

= (sin (a) + 1) (sin (a) - 1)

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

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

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

分解

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

1 - sin^{2} (a)

因式分解出通项

- 1

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

分解

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

sin^{2} (a) - 1

将

1

改写为

1^{2}

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

使用平方差公式:

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

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

= (sin (a) + 1) (sin (a) - 1)

= - (sin (a) + 1) (sin (a) - 1)

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

消掉

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

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

约分:

sin (a) + 1

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

约分:

sin (a) - 1

= sin^{2} (a) + 1

= sin^{2} (a) + 1

= sin^{2} (a) + 1

使用毕达哥拉斯恒等式:

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

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

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

化简

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

1 - cos^{2} (a) + 1

对同类项分组

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

数字相加:

1 + 1 = 2

= - cos^{2} (a) + 2

= - cos^{2} (a) + 2

= - cos^{2} (a) + 2

= 2 - cos^{2} (a)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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