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受欢迎的三角函数 >

证明 cos^4(t)-sin^4(t)=1-2sin^2(t)

解答

证明

cos^{4} (t) - sin^{4} (t) = 1 - 2 sin^{2} (t)

解答

真

求解步骤

cos^{4} (t) - sin^{4} (t) = 1 - 2 sin^{2} (t)

调整左侧

cos^{4} (t) - sin^{4} (t)

分解

cos^{4} (t) - sin^{4} (t) : (cos^{2} (t) + sin^{2} (t)) (cos (t) + sin (t)) (cos (t) - sin (t))

cos^{4} (t) - sin^{4} (t)

将

cos^{4} (t) - sin^{4} (t)

改写为

(cos^{2} (t))^{2} - (sin^{2} (t))^{2}

cos^{4} (t) - sin^{4} (t)

使用指数法则:

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

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

= cos^{4} (t) - (sin^{2} (t))^{2}

使用指数法则:

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

cos^{4} (t) = (cos^{2} (t))^{2}

= (cos^{2} (t))^{2} - (sin^{2} (t))^{2}

= (cos^{2} (t))^{2} - (sin^{2} (t))^{2}

使用平方差公式:

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

(cos^{2} (t))^{2} - (sin^{2} (t))^{2} = (cos^{2} (t) + sin^{2} (t)) (cos^{2} (t) - sin^{2} (t))

= (cos^{2} (t) + sin^{2} (t)) (cos^{2} (t) - sin^{2} (t))

分解

cos^{2} (t) - sin^{2} (t) : (cos (t) + sin (t)) (cos (t) - sin (t))

cos^{2} (t) - sin^{2} (t)

使用平方差公式:

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

cos^{2} (t) - sin^{2} (t) = (cos (t) + sin (t)) (cos (t) - sin (t))

= (cos (t) + sin (t)) (cos (t) - sin (t))

= (cos^{2} (t) + sin^{2} (t)) (cos (t) + sin (t)) (cos (t) - sin (t))

= (cos (t) + sin (t)) (cos (t) - sin (t)) (cos^{2} (t) + sin^{2} (t))

使用三角恒等式改写

(cos (t) + sin (t)) (cos (t) - sin (t)) (cos^{2} (t) + sin^{2} (t))

使用毕达哥拉斯恒等式:

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

= (cos (t) + sin (t)) (cos (t) - sin (t)) \cdot 1

化简

= (cos (t) + sin (t)) (cos (t) - sin (t))

乘开

(cos (t) + sin (t)) (cos (t) - sin (t)) : cos^{2} (t) - sin^{2} (t)

(cos (t) + sin (t)) (cos (t) - sin (t))

使用平方差公式:

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

a = cos (t), b = sin (t)

= cos^{2} (t) - sin^{2} (t)

= cos^{2} (t) - sin^{2} (t)

使用倍角公式:

cos^{2} (t) - sin^{2} (t) = cos (2 t)

= cos (2 t)

使用倍角公式:

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

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

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

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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