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受欢迎的三角函数 >

5sin(2θ)=7cos(2θ)

解答

5 sin (2 θ) = 7 cos (2 θ)

解答

θ = \frac{0.95054 \dots}{2} + \frac{πn}{2}

+1

度数

θ = 27.23116 \dots^{\circ} + 9 0^{\circ} n

求解步骤

5 sin (2 θ) = 7 cos (2 θ)

使用三角恒等式改写

5 sin (2 θ) = 7 cos (2 θ)

在两边除以

cos (2 θ), cos (2 θ) \neq = 0

\frac{5 sin ( 2 θ )}{cos ( 2 θ )} = \frac{7 cos ( 2 θ )}{cos ( 2 θ )}

化简

\frac{5 sin ( 2 θ )}{cos ( 2 θ )} = 7

使用基本三角恒等式:

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

5 tan (2 θ) = 7

5 tan (2 θ) = 7

两边除以

5

5 tan (2 θ) = 7

两边除以

5

\frac{5 tan ( 2 θ )}{5} = \frac{7}{5}

化简

tan (2 θ) = \frac{7}{5}

tan (2 θ) = \frac{7}{5}

使用反三角函数性质

tan (2 θ) = \frac{7}{5}

tan (2 θ) = \frac{7}{5}

的通解

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

2 θ = arctan (\frac{7}{5}) + πn

2 θ = arctan (\frac{7}{5}) + πn

解

2 θ = arctan (\frac{7}{5}) + πn : θ = \frac{arctan ( \frac{7}{5} )}{2} + \frac{πn}{2}

2 θ = arctan (\frac{7}{5}) + πn

两边除以

2

2 θ = arctan (\frac{7}{5}) + πn

两边除以

2

\frac{2 θ}{2} = \frac{arctan ( \frac{7}{5} )}{2} + \frac{πn}{2}

化简

θ = \frac{arctan ( \frac{7}{5} )}{2} + \frac{πn}{2}

θ = \frac{arctan ( \frac{7}{5} )}{2} + \frac{πn}{2}

θ = \frac{arctan ( \frac{7}{5} )}{2} + \frac{πn}{2}

以小数形式表示解

θ = \frac{0.95054 \dots}{2} + \frac{πn}{2}

作图

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