Question

Expansion of following functions are given below:
A. \[ \sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots \]
B. \[ \cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots \]
C. \[ \sin x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots \]
D. \[ \cos x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots \] Choose the correct answer from the options given below:

• A. A, D
• B. B, C
• C. A, C
• D. A, B

Hint:

\(\sin x\) contains odd powers of \(x\), while \(\cos x\) contains even powers of \(x\).

Answer 1

The Correct Option is D

Concept:
Maclaurin series expansions of \(\sin x\) and \(\cos x\) are standard results.

Step 1: Expansion of \(\sin x\). \[ \sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots \] So statement A is correct.

Step 2: Expansion of \(\cos x\). \[ \cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots \] So statement B is correct.

Step 3: Check C and D.
Statement C gives the expansion of \(\cos x\), not \(\sin x\).
Statement D gives the expansion of \(\sin x\), not \(\cos x\). \[ C,D \text{ are incorrect} \] \[ \therefore \text{Correct Answer is (D)} \]