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受欢迎的三角函数 >

4cos(2θ)=4sin(2θ)

解答

4 cos (2 θ) = 4 sin (2 θ)

解答

θ = \frac{π}{8} + \frac{πn}{2}

+1

度数

θ = 22. 5^{\circ} + 9 0^{\circ} n

求解步骤

4 cos (2 θ) = 4 sin (2 θ)

使用三角恒等式改写

4 cos (2 θ) = 4 sin (2 θ)

在两边除以

cos (2 θ), cos (2 θ) \neq = 0

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

化简

4 = \frac{4 sin ( 2 θ )}{cos ( 2 θ )}

使用基本三角恒等式:

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

4 = 4 tan (2 θ)

交换两边

4 tan (2 θ) = 4

4 tan (2 θ) = 4

两边除以

4

4 tan (2 θ) = 4

两边除以

4

\frac{4 tan ( 2 θ )}{4} = \frac{4}{4}

化简

tan (2 θ) = 1

tan (2 θ) = 1

tan (2 θ) = 1

的通解

tan (x)

周期表（周期为

πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

2 θ = \frac{π}{4} + πn

2 θ = \frac{π}{4} + πn

解

2 θ = \frac{π}{4} + πn : θ = \frac{π}{8} + \frac{πn}{2}

2 θ = \frac{π}{4} + πn

两边除以

2

2 θ = \frac{π}{4} + πn

两边除以

2

\frac{2 θ}{2} = \frac{\frac{π}{4}}{2} + \frac{πn}{2}

化简

\frac{2 θ}{2} = \frac{\frac{π}{4}}{2} + \frac{πn}{2}

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

\frac{\frac{π}{4}}{2} + \frac{πn}{2} : \frac{π}{8} + \frac{πn}{2}

\frac{\frac{π}{4}}{2} + \frac{πn}{2}

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

使用分式法则:

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

= \frac{π}{4 \cdot 2}

数字相乘:

4 \cdot 2 = 8

= \frac{π}{8}

= \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

作图

流行的例子

tan(x)=-sqrt(3),0<= x<= 2pi

tan (x) = - 3, 0 \leq x \leq 2 π

5 tan^{2} (A) - 7 = 8

tan (x) = \frac{28}{16}

2sin^2(x)=3-sin(x)

2 sin^{2} (x) = 3 - sin (x)

9tan(2x)-18cos(x)=0

9 tan (2 x) - 18 cos (x) = 0