Trigonometry Trigonometric Functions Given: Sin θ = (2/3), Find: Cos θ
1 answer:
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<h2>Answer with explanation:</h2>
We are asked to prove by the method of mathematical induction that:
where n is a positive integer.
- Let us take n=1
then we have:
Hence, the result is true for n=1.
- Let us assume that the result is true for n=k
i.e.
- Now, we have to prove the result for n=k+1
i.e.
<u>To prove:</u>
Let us take n=k+1
Hence, we have:
( Since, the result was true for n=k )
Hence, we have:
Also, we know that:
(
Since, for n=k+1 being a positive integer we have:
)
Hence, we have finally,
Hence, the result holds true for n=k+1
Hence, we may infer that the result is true for all n belonging to positive integer.
i.e.
where n is a positive integer.
Consider the function 
, which has derivative 
.
The linear approximation of
for some value 
within a neighborhood of 
is given by
Let
. Then 
can be estimated to be
![\sqrt[3]{63.97}\approx4-\dfrac{0.03}{48}=3.999375](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B63.97%7D%5Capprox4-%5Cdfrac%7B0.03%7D%7B48%7D%3D3.999375)
Since
for 
, it follows that must be strictly increasing over that part of its domain, which means the linear approximation lies strictly above the function . This means the estimated value is an overestimation.
Indeed, the actual value is closer to the number 3.999374902...
The linear approximation of
Let
Since
Indeed, the actual value is closer to the number 3.999374902...