Show that each fn is bounded on A, but the pointwise limit f of the sequence is not bounded on A. Does (fn) converge uniformly t
o f on A?
1 answer:
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Answer:
The Law of Cosine : cos C =
Step-by-step explanation:
See the figure to understand the proof :
Let A Triangle ABC with sides a,b,c,
Draw a perpendicular on base AC of height H meet at point D
Divide base length b as AD = x -b and CD = x
By Pythagoras Theorem
In Triangle BDC And In Triangle BDA
a² = h² + x² ( 1 ) c² = h² + (x-b)²
c² = h² + x² + b² - 2xb ...(. 2)
From above eq 1 and 2
c² = (a² - x²) + x² + b² - 2xb
or, c² = a² + b² - 2xb .....(3)
Again in ΔBDC
cos C =
Or, cos C =
∴ x= a cos C
Now put ht value of x in eq 3
I.e, c² = a² + b² - 2ab cos C
Hence , cos C = Proved Answer
