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受欢迎的三角函数 >

证明 2cos^2(A)-1=1-2sin^2(A)

解答

证明

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

解答

真

求解步骤

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

调整左侧

2 cos^{2} (A) - 1

使用三角恒等式改写

2 cos^{2} (A) - 1

使用毕达哥拉斯恒等式:

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

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

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

化简

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

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

乘开

2 (1 - sin^{2} (A)) : 2 - 2 sin^{2} (A)

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

使用分配律:

a (b - c) = ab - a c

a = 2, b = 1, c = sin^{2} (A)

= 2 \cdot 1 - 2 sin^{2} (A)

数字相乘:

2 \cdot 1 = 2

= 2 - 2 sin^{2} (A)

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

化简

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

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

对同类项分组

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

数字相加/相减:

2 - 1 = 1

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

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

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

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

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

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 sec^{4} (t) - tan^{4} (t) = 1 + 2 tan^{2} (t)

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