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受欢迎的三角函数 >

-4sin(2θ)=5cos(2θ)

解答

- 4 sin (2 θ) = 5 cos (2 θ)

解答

θ = - \frac{0.89605 \dots}{2} + \frac{πn}{2}

+1

度数

θ = - 25.67009 \dots^{\circ} + 9 0^{\circ} n

求解步骤

- 4 sin (2 θ) = 5 cos (2 θ)

使用三角恒等式改写

- 4 sin (2 θ) = 5 cos (2 θ)

在两边除以

cos (2 θ), cos (2 θ) \neq = 0

\frac{- 4 sin ( 2 θ )}{cos ( 2 θ )} = \frac{5 cos ( 2 θ )}{cos ( 2 θ )}

化简

- \frac{4 sin ( 2 θ )}{cos ( 2 θ )} = 5

使用基本三角恒等式:

\frac{sin ( x )}{cos ( x )} = tan (x)

- 4 tan (2 θ) = 5

- 4 tan (2 θ) = 5

两边除以

- 4

- 4 tan (2 θ) = 5

两边除以

- 4

\frac{- 4 tan ( 2 θ )}{- 4} = \frac{5}{- 4}

化简

tan (2 θ) = - \frac{5}{4}

tan (2 θ) = - \frac{5}{4}

使用反三角函数性质

tan (2 θ) = - \frac{5}{4}

tan (2 θ) = - \frac{5}{4}

的通解

tan (x) = - a \Rightarrow x = arctan (- a) + πn

2 θ = arctan (- \frac{5}{4}) + πn

2 θ = arctan (- \frac{5}{4}) + πn

解

2 θ = arctan (- \frac{5}{4}) + πn : θ = - \frac{arctan ( \frac{5}{4} )}{2} + \frac{πn}{2}

2 θ = arctan (- \frac{5}{4}) + πn

化简

arctan (- \frac{5}{4}) + πn : - arctan (\frac{5}{4}) + πn

arctan (- \frac{5}{4}) + πn

利用以下特性:

arctan (- x) = - arctan (x)

arctan (- \frac{5}{4}) = - arctan (\frac{5}{4})

= - arctan (\frac{5}{4}) + πn

2 θ = - arctan (\frac{5}{4}) + πn

两边除以

2

2 θ = - arctan (\frac{5}{4}) + πn

两边除以

2

\frac{2 θ}{2} = - \frac{arctan ( \frac{5}{4} )}{2} + \frac{πn}{2}

化简

θ = - \frac{arctan ( \frac{5}{4} )}{2} + \frac{πn}{2}

θ = - \frac{arctan ( \frac{5}{4} )}{2} + \frac{πn}{2}

θ = - \frac{arctan ( \frac{5}{4} )}{2} + \frac{πn}{2}

以小数形式表示解

θ = - \frac{0.89605 \dots}{2} + \frac{πn}{2}

作图

流行的例子

tan (A) = \frac{3}{4}

2 sin (3 x) + 1 = 0

sqrt(3)sin(θ)=cos(θ),0<= θ<= 2pi

3 sin (θ) = cos (θ), 0 \leq θ \leq 2 π

cos(2θ)=-(sqrt(2))/2

cos (2 θ) = - \frac{2}{2}

tan(θ/2+pi/3)=-1

tan (\frac{θ}{2} + \frac{π}{3}) = - 1