How do you find the maclaurin series expansion of #sin2x#?

Question

Isabella Ackerman · Accepted Answer

#sin2x=sum_(n=0)^oo(2*4^n)/((2n+1)!)x^(2n+1)#.

Any general Maclauren series for a function #f(x)# may be expressed as #f(x)=sum_(n=0)^oo(f^ (n) (x))/(n!)x^n# Thus, in this instance, if we extract the function's first few derivatives, we obtain #f=sin2x=>f(0)=0# #f'=2cos2x=>f'(0)=2# #f''=-4sin2x=>f''(0)=0# #f'''=8cos2x=>f'''(0)=8# #f^4=-16sin2x=>f^4(0)=0# #f^5=32cos2x=>f^5(0)=32# Continuing in this way, we eventually get that every even term is zero and hence vanishes from the Maclaren series, and every odd term has value #(2*4^n)#. As a result, the function's Mclauen power series expansion is: #sin2x=sum_(n=0)^oo(2*4^n)/((2n+1)!)x^(2n+1)#.

Isaiah Allen · Answer

undefined The Maclaurin series expansion of $ \sin(2x) $ can be found by using the Maclaurin series expansion of $ \sin(x) $ and substituting $2x$ for $x$ in the series. The Maclaurin series expansion of $ \sin(x) $ is $ x - \frac{{x^3}}{3!} + \frac{{x^5}}{5!} - \frac{{x^7}}{7!} + \cdots $. Substituting $2x$ for $x$ gives $ 2x - \frac{{(2x)^3}}{3!} + \frac{{(2x)^5}}{5!} - \frac{{(2x)^7}}{7!} + \cdots $. Simplifying this expression gives the Maclaurin series expansion of $ \sin(2x) $.