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受欢迎的三角函数 >

cos(2θ)=sin(2θ)

解答

cos (2 θ) = 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

cos (2 θ) = sin (2 θ)

使用三角恒等式改写

cos (2 θ) = sin (2 θ)

在两边除以

cos (2 θ), cos (2 θ) \neq = 0

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

化简

1 = \frac{sin ( 2 θ )}{cos ( 2 θ )}

使用基本三角恒等式:

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

1 = tan (2 θ)

交换两边

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

2cos(x/2)=sqrt(3)

2 cos (\frac{x}{2}) = 3

tan(x)(cot(x)-(sqrt(3))/3)=0

tan (x) (cot (x) - \frac{3}{3}) = 0

24sin^2(x)=24-12cos(x)

24 sin^{2} (x) = 24 - 12 cos (x)

sqrt(3)tan(x)=1

3 tan (x) = 1

2cos^2(t)+3cos(t)+1=0

2 cos^{2} (t) + 3 cos (t) + 1 = 0