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Trigonometry

Advanced Questions on Trignometry Part - 6

Author: Leonardodvin

Ques: 25 Show that one can use a composition of trigonometry buttons such as, $\sin,\cos,\tan,\sin^{-1},\cos^{-1},$ and $\tan^{-1,}$ to replace the broken reciprocal button on a calculator.

Solution: Because $\cos^{-1}\sin \theta=\pi/2-\theta$

and

$\tan (\pi/2-\theta)=1/\tan \theta$

for

$0<\theta <\pi/2,$

we have for any x >0,

$\tan \cos^{-1}\sin\tan^{-1}x=\tan \Big (\frac {\pi}{2}-\tan^{-1}x\Big )=\frac {1}{x},$

as desired. It is not difficult to check that $\tan\sin^{-1}\cos\tan^{-1}$ will also do the trick.

Ques: 26 Prove that in a triangle ABC,

$\frac {a-b}{a+b}=\tan \frac {A-B}{2}\tan \frac {C}{2}.$

Solution: From the law of sines and the sum-to-product formulas, we have

$\frac {a-b}{a+b}=\frac {\sin A-\sin B}{\sin A+\sin B}=\frac {2\sin \frac {A-B}{2}\cos \frac {A+B}{2}}{2\sin \frac {A+B}{2}\cos \frac {A-B}{2}}$

$=\tan \frac {A-B}{2}\cot \frac {A+B}{2}=\tan \frac {A-B}{2} \tan \frac {C}{2},$

as desired.

Ques: 27 Let a, b, c be real numbers, all different from âˆ'1 and 1, such that a +b+c = abc. Prove that

$\frac {a}{1-a^2}+\frac {b}{1-b^2}+\frac {c}{1-c^2}=\frac {4abc}{(1-a^2)(1-b^2)(1-c^2)}.$

Solution: Let $a=\tan x,b=\tan y,c=\tan z,$ where $x,y,z\not=\frac {k\pi}{4},$ for all integers k. The condition a + b + c = abc translates to tan(x + y + z) = 0, as indicated in notes after Question 13(1). From the double-angle formulas, it follows that

$\tan (2x+2y+2z)=\frac {2\tan (x+y+z)}{1-\tan^2(x+y+z)}=0.$

Hence

$\tan 2x+\tan 2y+\tan 2z=\tan 2x \tan 2y\tan 2z,$

using a similar argument to the one in Question 13(1). This implies that

$\frac {2\tan x}{1-\tan^2 x}+\frac {2\tan y}{1-\tan^2 y}+\frac {2\tan z}{1-\tan^2 z}$

$=\frac {2\tan x}{1-\tan^2 x}.\frac {2\tan y}{1-\tan^2 y}.\frac {2\tan z}{1-\tan^2 z}$

and the conclusion follows.

Ques: 28 Prove that a triangle ABC is isosceles if and only if

$a\cos B+b\cos C+c\cos A=\frac {a+b+c}{2}.$

Solution: By the extended law of sines, a = 2R sin A, b = 2R sin B, and c = 2R sin C. The desired identity is equivalent to

$2 \sin A\cos B+2 \sin B \cos C+2\sin C\cos A=\sin A+\sin B+\sin C,$

or

$\sin (A+B)+\sin (A-B)+\sin (B+C)$

$+\sin (B-C)+\sin (C+A)+\sin (C-A)$

$=\sin A+\sin B+\sin C.$

Because

A+B+C=180^0,

$\sin (A+B)=\sin C,$

$\sin(B+C)=\sin A,\sin (C+A)=\sin B.$

The last equality simplifies to

$\sin (A-B)+\sin (B-C)+\sin(C-A)=0,$

which in turn is equivalent to

$4\sin \frac {A-B}{2}\sin \frac {B-C}{2} \sin \frac {C-A}{2}=0,$

by Question 7. The conclusion now follows.

Ques: 29 Prove that $\cos 1^0$ is an irrational number.

Solution: Assume, for the sake of contradiction, that $\cos 1^0$ is rational. Then so is $\cos 2^0=2\cos^2 1^0-1.$ Using the identity

$\cos(n^0+1^0)+ \cos(n ^0 -1 ^0)= 2\cos n ^0 \cos 1 ^0, \qquad \qquad \qquad \qquad (*)$

we obtain by strong induction that $\cos n^0$ is rational for all integers $n\underline >1.$ But this is clearly false, because, for example, $\cos 30^0$ is not rational, yielding a contradiction.

Note: For the reader not familiar with the idea of induction. We can reason in the following way. Under the assumption that both $\cos 1^0$ and $\cos 2^0$ are rational, relation (âˆ-) implies that $\cos 3^0$ is rational, by setting n = 2 in the relation (âˆ-). Similarly, by the assumption that both $\cos 2^0$ and $\cos 3^0$ are rational, relation (âˆ-) implies that $\cos 4^0$ is rational, by setting n = 4 in the relation (âˆ-). And so on.We conclude that $\cos n^0$ is rational, for all positive integers n, under the assumption that $\cos 1^0$ is rational.

Ques: 30 Prove that $\frac {\sin^3 a}{\sin b}+\frac {\cos^3 a}{\cos b}\underline >\sec (a-b)$ for all $0<b,b<\frac {\pi}{2}.$

Solution: Multiplying the two sides of the inequality by

$\sin a \sin b+\cos a \cos b=\cos(a-b),$ we obtain the equivalent form

$\Big (\frac {\sin^3 a}{\sin b}+\frac {\cos^3 a}{\cos b}\Big ) (\sin a \sin b+\cos a\cos b)\underline >1.$

But this follows from Cauchy-Schwarz inequality because according to this inequality, the left-hand side is greater than or equal to $(\sin^2 a+\cos^2 a)^2=1.$