Math, asked by raghwendra122, 7 hours ago

expand six x in powers of (x - pie/2). hence find sin 91 degree to four places of decimals

Answers

Answered by tejrambhatt54
1

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

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