What is taylor sin(θ) ?

The answer to taylor sin(θ) is sum from n=0 to infinity}(-1)^n(x^{2n+1 of)/((2n+1)!)

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taylor sin(θ)

taylor

sin (θ)

n = 0 \sum \infty (- 1)^{n} \frac{x ^{2 n + 1}}{( 2 n + 1 ) !}

Solution steps

Apply the Taylor Formula

= 0 + \frac{\frac{d}{d θ} ( sin ( θ ) ) ( 0 )}{1 !} θ + \frac{\frac{d ^{2}}{d θ ^{2}} ( sin ( θ ) ) ( 0 )}{2 !} θ^{2} + \frac{\frac{d ^{3}}{d θ ^{3}} ( sin ( θ ) ) ( 0 )}{3 !} θ^{3} + \dots

Evaluate Derivatives

= 0 + \frac{1}{1 !} θ + \frac{0}{2 !} θ^{2} + \frac{- 1}{3 !} θ^{3} + \frac{0}{4 !} θ^{4} + \frac{1}{5 !} θ^{5} + \frac{0}{6 !} θ^{6} + \frac{- 1}{7 !} θ^{7} + \frac{0}{8 !} θ^{8} + \frac{1}{9 !} θ^{9} + \dots

Refine

= \frac{1}{1 !} θ - \frac{1}{3 !} θ^{3} + \frac{1}{5 !} θ^{5} - \frac{1}{7 !} θ^{7} + \frac{1}{9 !} θ^{9} + \dots

The series sum representation

= n = 0 \sum \infty (- 1)^{n} \frac{x ^{2 n + 1}}{( 2 n + 1 ) !}

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What is taylor sin(θ) ?
The answer to taylor sin(θ) is sum from n=0 to infinity}(-1)^n(x^{2n+1 of)/((2n+1)!)