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受欢迎的三角函数 >

solvefor θ,ρ=(4sin(φ)cos(θ))

解答

求解

θ, ρ = (4 sin (φ) cos (θ))

解答

θ = arccos (\frac{ρ}{4 sin ( φ )}) + 2 πn, θ = - arccos (\frac{ρ}{4 sin ( φ )}) + 2 πn

求解步骤

ρ = (4 sin (φ) cos (θ))

交换两边

4 sin (φ) cos (θ) = ρ

两边除以

4 sin (φ); φ \neq = 2 πn, φ \neq = π + 2 πn

4 sin (φ) cos (θ) = ρ

两边除以

4 sin (φ); φ \neq = 2 πn, φ \neq = π + 2 πn

\frac{4 sin ( φ ) cos ( θ )}{4 sin ( φ )} = \frac{ρ}{4 sin ( φ )}; φ \neq = 2 πn, φ \neq = π + 2 πn

化简

cos (θ) = \frac{ρ}{4 sin ( φ )}; φ \neq = 2 πn, φ \neq = π + 2 πn

cos (θ) = \frac{ρ}{4 sin ( φ )}; φ \neq = 2 πn, φ \neq = π + 2 πn

使用反三角函数性质

cos (θ) = \frac{ρ}{4 sin ( φ )}

cos (θ) = \frac{ρ}{4 sin ( φ )}

的通解

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

θ = arccos (\frac{ρ}{4 sin ( φ )}) + 2 πn, θ = - arccos (\frac{ρ}{4 sin ( φ )}) + 2 πn

θ = arccos (\frac{ρ}{4 sin ( φ )}) + 2 πn, θ = - arccos (\frac{ρ}{4 sin ( φ )}) + 2 πn

作图

流行的例子

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sin^{2} (a) - cos^{2} (a) = 1

arcsin (x) = - 1

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cos^{2} (\frac{x}{3}) - sin^{2} (\frac{x}{3}) = - \frac{1}{2}

sin (x) = 0.014

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