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受欢迎的三角函数 >

cos(2θ)=((sqrt(57))/(11))

解答

cos (2 θ) = (\frac{57}{11})

解答

θ = \frac{0.81433 \dots}{2} + πn, θ = π - \frac{0.81433 \dots}{2} + πn

+1

度数

θ = 23.32912 \dots^{\circ} + 18 0^{\circ} n, θ = 156.67087 \dots^{\circ} + 18 0^{\circ} n

求解步骤

cos (2 θ) = (\frac{57}{11})

使用反三角函数性质

cos (2 θ) = \frac{57}{11}

cos (2 θ) = \frac{57}{11}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

2 θ = arccos (\frac{57}{11}) + 2 πn, 2 θ = 2 π - arccos (\frac{57}{11}) + 2 πn

2 θ = arccos (\frac{57}{11}) + 2 πn, 2 θ = 2 π - arccos (\frac{57}{11}) + 2 πn

解

2 θ = arccos (\frac{57}{11}) + 2 πn : θ = \frac{arccos ( \frac{57}{11} )}{2} + πn

2 θ = arccos (\frac{57}{11}) + 2 πn

两边除以

2

2 θ = arccos (\frac{57}{11}) + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{arccos ( \frac{57}{11} )}{2} + \frac{2 πn}{2}

化简

θ = \frac{arccos ( \frac{57}{11} )}{2} + πn

θ = \frac{arccos ( \frac{57}{11} )}{2} + πn

解

2 θ = 2 π - arccos (\frac{57}{11}) + 2 πn : θ = π - \frac{arccos ( \frac{57}{11} )}{2} + πn

2 θ = 2 π - arccos (\frac{57}{11}) + 2 πn

两边除以

2

2 θ = 2 π - arccos (\frac{57}{11}) + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{2 π}{2} - \frac{arccos ( \frac{57}{11} )}{2} + \frac{2 πn}{2}

化简

θ = π - \frac{arccos ( \frac{57}{11} )}{2} + πn

θ = π - \frac{arccos ( \frac{57}{11} )}{2} + πn

θ = \frac{arccos ( \frac{57}{11} )}{2} + πn, θ = π - \frac{arccos ( \frac{57}{11} )}{2} + πn

以小数形式表示解

θ = \frac{0.81433 \dots}{2} + πn, θ = π - \frac{0.81433 \dots}{2} + πn

作图

流行的例子

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

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

solvefor θ,(cos(θ))/(sin(θ))=-pi/6

so l v e f or θ, \frac{cos ( θ )}{sin ( θ )} = - \frac{π}{6}

tan(θ)=(10)/(-5.32)

tan (θ) = \frac{10}{- 5.32}

2sin(2x)+cos(2x)=4cos(2x)

2 sin (2 x) + cos (2 x) = 4 cos (2 x)

sec(t)= 1/((sqrt(2))/2)

sec (t) = \frac{1}{\frac{2}{2}}