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受欢迎的三角函数 >

证明 sin^2(b)csc^2(b)-sin^2(b)=cos^2(b)

解答

证明

sin^{2} (b) csc^{2} (b) - sin^{2} (b) = cos^{2} (b)

解答

真

求解步骤

sin^{2} (b) csc^{2} (b) - sin^{2} (b) = cos^{2} (b)

调整左侧

sin^{2} (b) csc^{2} (b) - sin^{2} (b)

用 sin, cos 表示

- sin^{2} (b) + csc^{2} (b) sin^{2} (b)

使用基本三角恒等式:

csc (x) = \frac{1}{sin ( x )}

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

化简

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

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

(\frac{1}{sin ( b )})^{2} sin^{2} (b) = 1

(\frac{1}{sin ( b )})^{2} sin^{2} (b)

(\frac{1}{sin ( b )})^{2} = \frac{1}{sin ^{2} ( b )}

(\frac{1}{sin ( b )})^{2}

使用指数法则:

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

= \frac{1 ^{2}}{sin ^{2} ( b )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( b )}

= \frac{1}{sin ^{2} ( b )} sin^{2} (b)

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot sin ^{2} ( b )}{sin ^{2} ( b )}

约分:

sin^{2} (b)

= 1

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

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

= 1 - sin^{2} (b)

使用三角恒等式改写

1 - sin^{2} (b)

使用毕达哥拉斯恒等式:

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

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

= cos^{2} (b)

= cos^{2} (b)

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 cot(pi-x)=-cot(x)

p ro v e cot (π - x) = - cot (x)

证明 (csc(B))/(tan(B)+cot(B))=cos(B)

p ro v e \frac{csc ( B )}{tan ( B ) + cot ( B )} = cos (B)

证明 sin((17pi)/2+x)=cos(x)

p ro v e sin (\frac{17 π}{2} + x) = cos (x)

证明 csc(-x)-sin(-x)=cos(-x)cot(-x)

p ro v e csc (- x) - sin (- x) = cos (- x) cot (- x)

证明 sin(α)=(2tan(α/2))/(1+tan^2(α/2))

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