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受欢迎的三角函数 >

csc(2θ)= 1/2 (sec(θ))(cos(θ))

解答

csc (2 θ) = \frac{1}{2} (sec (θ)) (cos (θ))

解答

θ \in R 无解

求解步骤

csc (2 θ) = \frac{1}{2} (sec (θ)) (cos (θ))

两边减去

\frac{1}{2} sec (θ) cos (θ)

csc (2 θ) - \frac{1}{2} sec (θ) cos (θ) = 0

化简

csc (2 θ) - \frac{1}{2} sec (θ) cos (θ) : \frac{2 csc ( 2 θ ) - sec ( θ ) cos ( θ )}{2}

csc (2 θ) - \frac{1}{2} sec (θ) cos (θ)

\frac{1}{2} sec (θ) cos (θ) = \frac{sec ( θ ) cos ( θ )}{2}

\frac{1}{2} sec (θ) cos (θ)

分式相乘:

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

= \frac{1 \cdot sec ( θ ) cos ( θ )}{2}

乘以:

1 \cdot sec (θ) = sec (θ)

= \frac{sec ( θ ) cos ( θ )}{2}

= csc (2 θ) - \frac{sec ( θ ) cos ( θ )}{2}

将项转换为分式:

csc (2 θ) = \frac{csc ( 2 θ ) 2}{2}

= \frac{csc ( 2 θ ) \cdot 2}{2} - \frac{sec ( θ ) cos ( θ )}{2}

因为分母相等，所以合并分式:

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

= \frac{csc ( 2 θ ) \cdot 2 - sec ( θ ) cos ( θ )}{2}

\frac{2 csc ( 2 θ ) - sec ( θ ) cos ( θ )}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

2 csc (2 θ) - sec (θ) cos (θ) = 0

使用三角恒等式改写

2 csc (2 θ) - sec (θ) cos (θ)

sec (θ) cos (θ) = 1

sec (θ) cos (θ)

用 sin, cos 表示

sec (θ) cos (θ)

使用基本三角恒等式:

sec (θ) = \frac{1}{cos ( θ )}

= \frac{1}{cos ( θ )} cos (θ)

\frac{1}{cos ( θ )} cos (θ) = 1

\frac{1}{cos ( θ )} cos (θ)

分式相乘:

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

= \frac{1 cos ( θ )}{cos ( θ )}

约分:

cos (θ)

= 1

= 1

= 2 csc (2 θ) - 1

2 csc (2 θ) - 1 = 0

将

1

到右边

2 csc (2 θ) - 1 = 0

两边加上

1

2 csc (2 θ) - 1 + 1 = 0 + 1

化简

2 csc (2 θ) = 1

2 csc (2 θ) = 1

两边除以

2

2 csc (2 θ) = 1

两边除以

2

\frac{2 csc ( 2 θ )}{2} = \frac{1}{2}

化简

csc (2 θ) = \frac{1}{2}

csc (2 θ) = \frac{1}{2}

csc (x) \leq - 1 or csc (x) \geq 1

θ \in R 无解

作图

流行的例子

sec^2(x)+2sec(x)-8=0

sec^{2} (x) + 2 sec (x) - 8 = 0

3cos^2(x)+4sin(x)-4=0

3 cos^{2} (x) + 4 sin (x) - 4 = 0

sin (x) = \frac{7}{15}

-4sin(x)=-cos^2(x)+4,0<= x<= 2pi

- 4 sin (x) = - cos^{2} (x) + 4, 0 \leq x \leq 2 π

sin(2x)cot(x)=0

sin (2 x) cot (x) = 0