Answer:
Cos x = 1 -
+
-
+ ...
Step-by-step explanation:
We use Taylor series expansion to answer this question.
We have to find the expansion of cos x at x = 0
f(x) = cos x, f'(x) = -sin x, f''(x) = -cos x, f'''(x) = sin x, f''''(x) = cos x
Now we evaluate them at x = 0.
f(0) = 1, f'(0) = 0, f''(0) = -1, f'''(0) = 0, f''''(0) = 1
Now, by Taylor series expansion we have
f(x) = f(a) + f'(a)(x-a) +
+
+
+ ...
Putting a = 0 and all the values from above in the expansion, we get,
Cos x = 1 -
+
-
+ ...
N = d - 5
2n/(d + 16) = n/d - 1/3 n/d
2n/(d + 16) = 2/3 n/d Divide by 2n
1 / (d + 16) = 1/3 d
Here's the tricky part.
d + 16 = 3d the two denominators are equal. the numerators are both 1.
16 = 2d
d = 8
so the numerator is d - 5
n = 8 - 5
n = 3
Let's see if it checks out.
n = 3
d = 8
2*3 = 6
8 + 16 = 24
New fraction 6/24 = 1/4
(3/8 - 1/3 ) = 1/8
3/8 - 1/8 = 1/4 so it checks with the original conditions put on it.
Answer:
4/-5
Step-by-step explanation:
Need more details than that