Advanced Questions on Trignometry Part - 7

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Advanced Questions on Trignometry Part - 7

Author: Leonardodvin

Ques: 31 If $\sin \alpha \cos \beta=-\frac {1}{2},$ what are the possible values of $\cos \alpha \sin \beta$ ?

Solution: Note that

$\sin(\alpha +\beta)=\sin \alpha \cos \beta$ $+\cos \alpha \sin \beta$ $=- \frac {1}{2}+$ $\cos \alpha \sin \beta.$

Because $-1<\sin (\alpha +\beta)\underline <1,$ it follows that $-\frac {1}{2}\underline <$ $\cos \alpha \sin \beta < \frac {3}{2}.$ Similarly, because $\sin (\alpha -\beta )=$ $\sin \alpha \cos$ $\beta -\cos \alpha$ $\sin \beta ,$ we conclude that $-\frac {3}{2}\underline<\cos$ $\alpha \sin \beta$ $< \frac {1}{2}.$ Combining the above results shows that

$-\frac {1}{2}\underline <\cos \alpha \sin \beta \underline < \frac {1}{2}.$

But we have not shown that indeed, $\cos \alpha \sin \beta$ can obtain all values in the interval $\Big [-\frac {1}{2},\frac {1}{2}\Big ].$ To do this, we consider

$(\cos \alpha \sin\beta)^2=$ $(1-\sin ^2 \alpha)(1-\cos ^2 \beta)$

$=1-(\sin^2 \alpha +\cos^2 \beta )+\sin^2 \alpha \cos^2 \beta$

$=\frac {5}{4}-(\sin^2 \alpha +\cos^2 \beta)$

$=\frac {5}{4}-(\sin \alpha +\cos \beta )^2$ $+2\sin \alpha \cos \beta$

$=\frac {1}{4}-(\sin \alpha +\cos \beta)^2.$

Let $x=\sin \alpha$ and $y=\cos \beta.$ Then $-1\underline <x,y,\underline <1$ and $xy=-\frac {1}{2}.$ Consider the range of the sum $s=\sin \alpha +\cos \beta =x+y.$ If $xy=-\frac {1}{2}$ and x+y=s, then x and y are the roots of the quadratic equation

$u^2-su-\frac {1}{2}=0. \qquad \qquad \qquad \qquad \qquad (*)$

Thus,

$\{x,y\}=\Big \{\frac {s+\sqrt {s^2+2}}{2},\frac {s-\sqrt {s^2+2}}{2}\Big \}.$

By checking the boundary condition

$\frac {s+\sqrt {s^2+2}}{2} \underline <1,$ we obtain $s\underline < \frac {1}{2}.$ By checking similar boundary conditions, we conclude that the equation (âˆ-) has a pair of solutions x and y with $-1\underline < x,y \underline < 1$ for all $-\frac {1}{2}$ $\underline < s \underline < \frac {1}{2}.$ Because both the sine and cosine functions are surjective functions from R to the interval [-1,1], the range of $s =\sin \alpha +\cos \beta$ is $[-\frac {1}{2}, \frac {1}{2}]$ for $\sin \alpha \cos \beta$ $=-\frac {1}{2}$ . Thus, the range of s^2 is $[0,\frac {1}{2}]$ . Thus the range of $(\cos \alpha \sin \beta ) ^2$ is $[0,\frac {1}{4}]$ , and so the reange of $\cos \alpha \sin \beta$ is $[-\frac {1}{2},\frac {1}{2}].$

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

$(ab+bc+ca-1)^2\underline < (a^2+1)(b^2+1)(c^2+1)$

Solution: Let a = tan x, b = tan y, c = tan z with $-\frac {\pi}{2}<x,y,z<\frac {\pi}{2}.$

Then $a^2+1=\sec^2 x,b^2+1=\sec^2 y,$ and $c^2+1=\sec^2 z.$

Multiplying by $\cos^2 x \cos^2 y \cos^2 z$ on both sides of the desired inequality gives

$[(ab+bc+ca-1)\cos x\cos y\cos z]^2\underline <1.$

Note that

$(ab+bc)\cos x\cos y\cos z=\sin x\sin y\cos z+\sin y\sin z\cos x$

$=\sin y\sin(x+z)$

and

$(ca-1)\cos x\cos y\cos z=\sin z\sin x\cos y-\cos x\cos y\cos z$

$=-\cos y\cos(x+z).$

Consequently, we obtain

$[(ab+bc+ca-1)\cos x\cos y\cos z]^2$

$=[\sin y\sin(x+z)-\cos y\cos(x+z)]^2$

$=\cos^2(x+y+z)\underline <1,$

as desired.

Ques: 33 Prove that

$(\sin x+a\cos x)(\sin x+b\cos x)\underline < 1+ \Big (\frac {a+b}{2}\Big )^2.$

Solution: If cos x = 0, the desired inequality reduces to $\sin^2 x \underline <1+\Big (\frac {a+b}{2}\Big )^2,$ which is clearly true. We assume that $\cos x\not=0.$ Dividing both sides of the desired inequality by $\cos^2 x$ gives

$(\tan x+a)(\tan x+b)\underline < \Big [1+\Big (\frac {a+b}{2}\Big )^2 \Big]\sec^2x.$

Set t = tan x. Then $\sec^2 x=1+t^2.$ The above inequality reduces to

$t^2+(a+b)t+ab\underline < \Big (\frac {a+b}{2}\Big )^2t^2+t^2+\Big (\frac {a+b}{2}\Big )^2+1,$

or

$\Big (\frac {a+b}{2}\Big )^2t^2$ $+1-(a+b)t+\Big (\frac {a+b}{2}\Big )^2$ $-ab \underline >0.$

The last inequality is equivalent to

$\Big (\frac {(a+b)t}{2}-1\Big )^2+\Big (\frac {a-b}{2}\Big )^2\underline >0,$

which is evident.

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

$|\sin a_1| + |\cos a_1|\underline > \sin ^2 a_1+ \cos^2 a_1=1.$

For the inductive step, in order to prove that

$|\sin a_1|+|\sin a_2|+\cdots + |\sin a_{n+1}| + |\cos(a_1+a_2 + \cdots + a_{n+1})|\underline >1,$

it suffices to show that

$|\sin a_{n+1}|+ |\cos(a_1 +$ $a_2 +\cdots +a_{n+1})|\underline >$ $|\cos(a_1+a_2+\cdots +a_n)|$

for all real numbers $a_1+a_2.......a_{n+1}$ Let s_k = a_1 +a_2+...+a_k For k = 1,2,.....n+1. The last inequality becomes $|\sin a_{n+1}|+|\cos s_{n+1}|$ $\underline >|\cos s_n|.$ Indeed by the addition and subtraction formulas, we have

$|\cos s_n|=|\cos(s_{n+1}-a_{n+1})|$

$=|\cos s_{n+1} \cos a_{n+1}+\sin s_{n+1}\sin a_{n+1}|$

$= |\cos s_{n+1} \cos a_{n+1}| + |\sin s_{n+1} \sin a_{n+1}|$

$\underline <|\cos s_{n+1}| + |\sin a_{n+1}|,$

as desired

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

$S=\{\sin \alpha , \sin 2 \alpha , \sin 3\alpha\}$

is equal to the set

$T=\{\cos \alpha , \cos 2 \alpha , \cos 3\alpha\}$ .

Solution: The answers are $\alpha= \frac{ \pi}{8}+\frac {k \pi}{2}$ for all intergers k

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

$\sin \alpha + \sin 2 \alpha +\sin 3 \alpha =\cos \alpha + \cos 2 \alpha +\cos 3 \alpha.$

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

$2 \sin 2 \alpha \cos \alpha+ \sin 2\alpha=2 \cos 2\alpha \cos \alpha + \cos 2\alpha$

or

$\sin 2 \alpha (2 \cos \alpha +1)=\cos 2 \alpha (2 \cos \alpha +1).$

if $2 \cos \alpha +1 = 0$ then $\cos \alpha = - \frac{1}{2}$ , and so $\alpha = \pm \frac{ 2 \pi}{3}+2k \pi$ for all intergers k. It is then not difficult to check that $S \not= T$ and both of S and T are not three-elements sets.

It follows that $2 \cos \alpha + 1 \not= 0$ , implying that $\sin 2 \alpha = \cos 2 \alpha$ ;that is, $\tan 2 \alpha = 1$ . The possible answers are $\alpha = \frac{ \pi}{8} + \frac{k \pi}{2}$ for all intergers k. Because $\frac{ \pi}{8} + \frac{3 \pi}{8} = \frac{ \pi}{2} , \cos \frac{\pi}{8} = \sin \frac{3 \pi}{8}.$ It not difficult to check that all such angles satisfy the conditions of the problem.

Ques: 36 Let $\{ T_n(x)\}^{\infty}_{n=0}$ be the sequence of polynomials such that $T_0(x)=1, T_1(x)= x,T_{i+1}=2xT_i(x)-T_{i-1}(x)$ for all positive integers i. The polynomial T_n(x) is called the nth Chebyshev polynomial.

(a) Prove that $T_{2n+1}(x)$ and $T_{2n}(x)$ are odd and even functions,respectively;

(b) Prove that $T_{n+1}(x) > T_n(x) > 1$ for real numbers x with x > 1;

(d) Determine all the roots of T_n(x) ;

(e) Determine all the roots of P_n(x)-T_n(x)-1.

Solution: Parts (a) and (b) are simple facts that will be useful in establishing (e).We present them together.
(a). We apply strong induction on n. Note that T_0 = 1 and T_1 = x are even and odd, respectively. Assume that $T_{2n-1}$ and $T_{2n}$ are odd and even, respectively. Then $2xT_{2n}$ is odd, and so $T_{2n+1} = 2xT_{2n}-T_{2n-1}$ is odd. Thus $2xT_{2n+1}$ is even, and so $T_{2n+2} = 2xT_{2n+1} - T_{2n}$ is even. This completes our induction.

(b). We apply strong induction on n.For n = 0, T_1(x) = x > 1 = T_0(x) for x > 1. Assume that $T_{n+1}(x) >$ T_n(x) > 1 for x > 1 and $n \underline < k,$ where k is some nonnegative integer. For n = k + 1, the induction hypothesis yields

$T_{k+2}(x) = 2xT_{k+1}(x)$ $- T_k(x) > 2T_{k+1}(x) - T_k(x)$

$= T_{k+1}(x) + T_{k+1}(x)$ $- T_k(x) > T_{k+1}(x),$

completing our induction.