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受欢迎的三角函数 >

证明 (cos(b))/(sec(b))+(sin(b))/(csc(b))=csc^2(b)-cot^2(b)

解答

证明

\frac{cos ( b )}{sec ( b )} + \frac{sin ( b )}{csc ( b )} = csc^{2} (b) - cot^{2} (b)

解答

真

求解步骤

\frac{cos ( b )}{sec ( b )} + \frac{sin ( b )}{csc ( b )} = csc^{2} (b) - cot^{2} (b)

调整左侧

\frac{cos ( b )}{sec ( b )} + \frac{sin ( b )}{csc ( b )}

用 sin, cos 表示

\frac{cos ( b )}{sec ( b )} + \frac{sin ( b )}{csc ( b )}

使用基本三角恒等式:

sec (x) = \frac{1}{cos ( x )}

= \frac{cos ( b )}{\frac{1}{c o s ( b )}} + \frac{sin ( b )}{csc ( b )}

使用基本三角恒等式:

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

= \frac{cos ( b )}{\frac{1}{c o s ( b )}} + \frac{sin ( b )}{\frac{1}{s i n ( b )}}

化简

\frac{cos ( b )}{\frac{1}{c o s ( b )}} + \frac{sin ( b )}{\frac{1}{s i n ( b )}} : cos^{2} (b) + sin^{2} (b)

\frac{cos ( b )}{\frac{1}{c o s ( b )}} + \frac{sin ( b )}{\frac{1}{s i n ( b )}}

\frac{cos ( b )}{\frac{1}{c o s ( b )}} = cos^{2} (b)

\frac{cos ( b )}{\frac{1}{c o s ( b )}}

使用分式法则:

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

= \frac{cos ( b ) cos ( b )}{1}

使用法则

\frac{a}{1} = a

= cos (b) cos (b)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

cos (b) cos (b) = cos^{1 + 1} (b)

= cos^{1 + 1} (b)

数字相加:

1 + 1 = 2

= cos^{2} (b)

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

\frac{sin ( b )}{\frac{1}{s i n ( b )}}

使用分式法则:

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

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

使用法则

\frac{a}{1} = a

= sin (b) sin (b)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

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

= sin^{1 + 1} (b)

数字相加:

1 + 1 = 2

= sin^{2} (b)

= cos^{2} (b) + sin^{2} (b)

= cos^{2} (b) + sin^{2} (b)

= cos^{2} (b) + sin^{2} (b)

使用三角恒等式改写

cos^{2} (b) + sin^{2} (b)

使用毕达哥拉斯恒等式:

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

= 1

使用毕达哥拉斯恒等式:

1 + cot^{2} (x) = csc^{2} (x)

1 = csc^{2} (x) - cot^{2} (x)

= csc^{2} (b) - cot^{2} (b)

= csc^{2} (b) - cot^{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 1+cot^2(-x)=csc^2(x)

p ro v e 1 + cot^{2} (- x) = csc^{2} (x)

证明 cos(θ+pi/4)=(sqrt(2))/2 (cos(θ)-sin(θ))

p ro v e cos (θ + \frac{π}{4}) = \frac{2}{2} (cos (θ) - sin (θ))

证明 5cos^2(x)+7sin^2(x)-5=2sin^2(x)

p ro v e 5 cos^{2} (x) + 7 sin^{2} (x) - 5 = 2 sin^{2} (x)

证明 csc(2x)=(sec(x)csc(x))/2

p ro v e csc (2 x) = \frac{sec ( x ) csc ( x )}{2}

证明 sec(2x)=((sec^2(x)))/(2-sec^2(x))

p ro v e sec (2 x) = \frac{( sec ^{2} ( x ) )}{2 - sec ^{2} ( x )}