How do you graph $r = 1 + sin (θ)$ $r = 1 + sin(theta)$ ?

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Answer 1 · 2018-06-18T04:09:05

As below.

$r = 1 + sin θ$ $r = 1+ sin theta$

Standard form of a sine function is $y = A sin (B x - C) + D$ $y = A sin (Bx - C) + D$

$A = 1, B = 1, C = 0, D = 1$ $A = 1, B = 1, C = 0, D = 1$

$Amplitude = | A | = 1$ $"Amplitude " = |A| = 1$

$Period = \frac{2 π}{| B |} = 2 π$ $"Period " = (2pi) / |B| = 2pi$

$Phase Shift = - \frac{C}{B} = 0$ $"Phase Shift " = -C / B = 0$

$Vertical Shift = D = 1$ $"Vertical Shift " = D = 1$

graph{sin (x) + 1 [-10, 10, -5, 5]}

How do you graph r=1+sin(θ)r = 1 + sin(theta)?