expand six x in powers of (x - pie/2). hence find sin 91 degree to four places of decimals
Answers
Answer:
The expansion of sin(x)sin(x) in powers of (x-\frac{\pi}{2})(x−
2
π
) using Taylor's series is
sin(x) = \sum_{k \in \mathbb{N}}\frac{(x-\frac{\pi}{2})^{4k}}{(4k)!}-\frac{(x-\frac{\pi}{2})^{4k+2}}{(4k+2)!}sin(x)=∑
k∈N
(4k)!
(x−
2
π
)
4k
−
(4k+2)!
(x−
2
π
)
4k+2
Step-by-step explanation:
By Taylor's Theorem,
sin(x) = sin(\frac{\pi}{2})+sin'(\frac{\pi}{2})(x-\frac{\pi}{2})+ \dots + \frac{sin^{(k)}(\frac{\pi}{2})}{k!}(x-\frac{\pi}{2})^ksin(x)=sin(
2
π
)+sin
′
(
2
π
)(x−
2
π
)+⋯+
k!
sin
(k)
(
2
π
)
(x−
2
π
)
k
But
sin^{(n)}(\frac{\pi}{2}) = sin(\frac{\pi}{2}) = 1sin
(n)
(
2
π
)=sin(
2
π
)=1 if n = 4kn=4k
sin^{(n)}(\frac{\pi}{2}) = -sin(\frac{\pi}{2}) = -1sin
(n)
(
2
π
)=−sin(
2
π
)=−1 if n = 4k+2n=4k+2
and
sin^{(n)}(\frac{\pi}{2}) = \pm cos(\frac{\pi}{2}) = 0sin
(n)
(
2
π
)=±cos(
2
π
)=0 otherwise
Therefore
sin(x) = \sum_{k \in \mathbb{N}}\frac{(x-\frac{\pi}{2})^{4k}}{(4k)!}-\frac{(x-\frac{\pi}{2})^{4k+2}}{(4k+2)!}sin(x)=∑
k∈N
(4k)!
(x−
2
π
)
4k
−
(4k+2)!
(x−
2
π
)
4k+2