How do you find the Maclaurin Series for #(sinx)(cosx)#?

Answer 1

#sin(x)cos(x)=1/2sum_(k=0)^oo(-1)^k(2x)^(2k+1)/((2k+1)!)#

We know that #sin(2x)=2sin(x)cos(x)# so
#sin(x)cos(x)=1/2sin(2x)# or
#sin(x)cos(x)=1/2sum_(k=0)^oo(-1)^k(2x)^(2k+1)/((2k+1)!)#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To find the Maclaurin series for ( \sin(x) \cdot \cos(x) ), we can use the Maclaurin series expansions of ( \sin(x) ) and ( \cos(x) ) and multiply them together.

The Maclaurin series expansions for ( \sin(x) ) and ( \cos(x) ) are:

[ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots ]

[ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots ]

To find the Maclaurin series for ( \sin(x) \cdot \cos(x) ), multiply the corresponding terms of the series expansions:

[ \sin(x) \cdot \cos(x) = \left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots\right) \times \left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots\right) ]

Now, multiply the terms to find the series expansion for the product:

[ \sin(x) \cdot \cos(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots - \frac{x^3}{2!} + \frac{x^5}{4!} - \frac{x^7}{6!} + \cdots ]

[ = x - \frac{x^3}{2!} - \frac{x^3}{3!} + \frac{x^5}{4!} + \frac{x^5}{5!} - \frac{x^7}{6!} - \frac{x^7}{7!} + \cdots ]

[ = x - \frac{x^3}{2} - \frac{x^3}{6} + \frac{x^5}{24} + \frac{x^5}{120} - \frac{x^7}{720} - \frac{x^7}{5040} + \cdots ]

[ = x - \frac{x^3}{2} + \frac{x^5}{12} - \frac{x^7}{90} + \cdots ]

So, the Maclaurin series for ( \sin(x) \cdot \cos(x) ) is:

[ \sin(x) \cdot \cos(x) = x - \frac{x^3}{2} + \frac{x^5}{12} - \frac{x^7}{90} + \cdots ]

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7