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受欢迎的三角函数 >

证明 sin^2(2θ)=(1-cos(4θ))/2

解答

证明

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

解答

真

求解步骤

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

调整右侧

\frac{1 - cos ( 4 θ )}{2}

使用三角恒等式改写

\frac{1 - cos ( 4 θ )}{2}

= \frac{1 - cos ( 2 \cdot 2 θ )}{2}

使用倍角公式:

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

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

化简

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

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

乘开

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

1 - (1 - 2 sin^{2} (2 θ))

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

- (1 - 2 sin^{2} (2 θ))

打开括号

= - (1) - (- 2 sin^{2} (2 θ))

使用加减运算法则

- (- a) = a, - (a) = - a

= - 1 + 2 sin^{2} (2 θ)

= 1 - 1 + 2 sin^{2} (2 θ)

1 - 1 = 0

= 2 sin^{2} (2 θ)

= \frac{2 sin ^{2} ( 2 θ )}{2}

数字相除:

\frac{2}{2} = 1

= sin^{2} (2 θ)

= sin^{2} (2 θ)

= sin^{2} (2 θ)

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 \frac{1}{tan ( x )} = arctan (x)

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p ro v e sec (- x) = \frac{1}{cos ( x )}

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p ro v e \frac{( sec ( θ ) - tan ( θ ) ) ^{2} + 1}{csc ( θ ) ( sec ( θ ) - tan ( θ ) )} = 2 tan (θ)

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p ro v e tan (\frac{x}{2}) = \frac{sin ( x )}{cos ( x ) + 1}