# Advanced Questions on Trignometry Part - 7

by Satish

Ques: 31 If what are the possible values of ?

Solution: %{font-family:verdana}Note that

Because it follows that Similarly, because we conclude that Combining the above results shows that

But we have not shown that indeed, can obtain all values in the interval To do this, we consider

Let and Then and Consider the range of the sum If and then x and y are the roots of the quadratic equation

Thus,

By checking the boundary condition

we obtain By checking similar boundary conditions, we conclude that the equation (âˆ-) has a pair of solutions x and y with for all Because both the sine and cosine functions are surjective functions from R to the interval [-1,1], the range of is for . Thus, the range of is . Thus the range of is , and so the reange of is %

Ques: 32 Let a, b, c be real numbers. Prove that

Solution: Let a = tan x, b = tan y, c = tan z with

Then and

Multiplying by on both sides of the desired inequality gives

Note that

and

Consequently, we obtain

as desired.

Ques: 33 Prove that

Solution: If cos x = 0, the desired inequality reduces to which is clearly true. We assume that Dividing both sides of the desired inequality by gives

Set t = tan x. Then The above inequality reduces to

or

The last inequality is equivalent to

which is evident.

Ques: 34 Prove that

Solution: We proceed by induction on n. The base case holds, because

For the inductive step, in order to prove that

it suffices to show that

for all real numbers Let For k = 1,2,…..n+1. The last inequality becomes Indeed by the addition and subtraction formulas, we have

as desired

Ques: 35 [Russia 2003, by Nazar Agakhanov] Find all angles Î± for which the three element set

is equal to the set

.

for all intergers k%

Because S = T, the sums of the elements in S and T are equal to each other that is,

Applying the sum-to-product formulas to the first and the third summands on each side of the last equation gives

or

%{font-family:verdana}if then , and so for all intergers k. It is then not difficult to check that and both of S and T are not three-elements sets.%

%{font-family:verdana}It follows that

, implying that ;that is,. The possible answers are for all intergers k. Because It not difficult to check that all such angles satisfy the conditions of the problem.%

Ques: 36 %{font-family:verdana}Let

be the sequence of polynomials such that for all positive integers i. The polynomial is called the nth Chebyshev polynomial.%

%{font-family:verdana}(a) Prove that

and are odd and even functions,respectively;%

%{font-family:verdana}(b) Prove that for real numbers x with x > 1;%

%{font-family:verdana}© Prove that for all nonegative intergers n;%

%{font-family:verdana}(d) Determine all the roots of ;%

%{font-family:verdana}(e) Determine all the roots of %

Solution: Parts (a) and (b) are simple facts that will be useful in establishing (e).We present them together.
%{font-family:verdana}(a). We apply strong induction on n. Note that

and are even and odd, respectively. Assume that and are odd and even, respectively. Then is odd, and so is odd. Thus is even, and so is even. This completes our induction.%

%{font-family:verdana}(b). We apply strong induction on n.For for x > 1. Assume that for x > 1 and where k is some nonnegative integer. For n = k + 1, the induction hypothesis yields

completing our induction.%

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GananathanFri, 30 Apr 2010 13:30:16 -0000

Basics of trigonometry the way you explained is good