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IIT-JEE 2009

Advanced Questions on Trignometry Part - 5

Author: Leonardodvin

Ques: 21 Let ABC be a triangle. Prove that

$(a) \quad \cos A\cos B \cos C \underline < \frac {1}{8};$

$(b) \quad \sin A\sin B \sin C \underline <\frac {3\sqrt {3}}{8};$

$(c) \quad \sin A+\sin B+\sin C \underline < \frac {3\sqrt {3}}{2};$

$(d) \quad \cos^2 A+\cos^2 B+\cos^2 C \underline >\frac {3}{4};$

$(e) \quad \sin^2 A+\sin^2 B+\sin^2 C\underline < \frac {9}{4};$

$(f) \quad \cos 2A+\cos 2B+\cos 2C\underline > -\frac {3}{2};$

$(g) \quad \sin 2A+\sin 2B+\sin 2C\underline < \frac {3\sqrt {3}}{2}.$

Solution: For part (a), if triangle ABC is non-acute, the left-hand side of the inequality is nonpositive, and so the inequality is clearly true.

If ABC is acute, then $\cos A,\cos B,\cos C$ are all positive. To establish (a) and (d), we need only note that the relation between (a) and (d) and Question 17 (d) is similar to that of Question 16 (a) and (b) and Question 15. (Please see the note after the solution of Question 16.)

The two inequalities in parts (d) and (e) are equivalent because $\cos2 x+\sin^2 x=1.$

By (e) and by the arithmetic-geometric means inequality, we have

$\frac {9}{4}\underline > \sin^2 A+\sin^2 B+\sin^2 C\underline >3 \sqrt [3]{\sin^2 A \sin^2 B\sin^2 C},$

from which (b) follows.

From $(a-b)^2+(b-c)^2+(c-a)^2\underline >0$ or by application of Cauchy-Schwarz inequality, we can show that $3(a^2+b^2+c^2)\underline >(a+b+c)^2.$ By (e) and by setting $a=\sin A,b=\sin B,c=\sin C,$ we obtain (c.)

Part (f) follows from (e) and $\cos 2x=2\cos^2 x-1.$ Finally, (g) follows from (b) and the identity

$\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C$

proved in Question 18(e).

Ques: 22 Prove that

$\frac {\tan 3x}{\tan x}=\tan \Big (\frac {\pi}{3}-x\Big )\tan \Big (\frac {\pi}{3}+x\Big )$

for all $x\not=\frac {k\pi}{6},$ where k is in Z.

Solution: From the triple-angle formulas, we have

$\tan 3x=\frac {3\tan x-\tan^3 x}{1-3 \tan^2 x}$

$=\tan x.\frac {(\sqrt {3}-\tan x)(\sqrt {3}+\tan x)}{(1-\sqrt {3}\tan x)(1+\sqrt {3}\tan x)}$

$=\frac {\sqrt {3}-\tan x}{1+\sqrt {3}\tan x}.\tan x.\frac {\sqrt {3}+\tan x}{1-\sqrt {3}\tan x}$

$=\tan \Big (\frac {\pi}{3}-x \Big ) \tan x \tan \Big (\frac {\pi}{3}+x \Big )$

for all $x\not=\frac {k\pi}{6},$ where k is in Z.

Ques: 23 [AMC12P 2002] Given that

$(1+\tan 1^0)(1+\tan 2^0)\cdots (1+\tan 45^0)=2^n,$

find n.

First Solution: Note that

$1+\tan k^0=1+\frac {\sin k^0}{\cos k^0}=\frac {\cos k^0+\sin k^0}{\cos k^0}$

$=\frac {\sqrt {2}\sin(45+k)^0}{\cos k^0}=\frac {\sqrt {2}\cos(45-k)^0}{\cos k^0}.$

Hence

$(1+\tan k^0)(1+\tan(45-k)^0)=\frac {\sqrt {2}\cos(45-k)^0}{\cos k^0}.\frac {\sqrt {2}\cos(k)^0}{\cos(45-k)^0}=2.$

It follows that

$(1+\tan 1^0)(1+\tan 2^0)\cdots (1+\tan 45^0)$

$=(1+\tan 1^0)(1+\tan 44^0)(1+\tan 2^0)(1+\tan 43^0)$

$\cdots (1+\tan 22^0)(1+\tan 23^0)(1+\tan 45^0)=2^{23}$

implying that n = 23.

Second Solution: Note that

$(1+\tan k^0)(1+\tan(45-k)^0)$

$=1+[\tan k^0+\tan(45-k)^0]+\tan k^0 \tan(45-k)^0$

$=1+\tan 45^0 [1-\tan k^0 \tan(45-k)^0]+\tan k^0 \tan(45-k)^0=2$

Hence

$(1+\tan 1^0)(1+\tan 2^0)\cdots (1+\tan 45^0)$

$=(1+\tan 1^0)(1+\tan 44^0)(1+\tan 2^0)(1+\tan 43^0)$

$\cdots (1+\tan 22^0)(1+\tan 23^0)(1+\tan 45^0)=2^{23},$

implying that n = 23.

Ques: 24 [AIME 2003] Let A=(0,0) and B=(b,2) be points in the coordinate plane. Let ABCDEF be a convex equilateral hexagon such that $\angle FAB=120^0,AB||DE,BC||EF,$ and CD||FA, and the y coordinates of its vertices are distinct elements of the set $\{0,2,4,6,8,10 \}.$ The area of the hexagon can be written in the form $m\sqrt {n},$ where m and n are positive integers and n is not divisible by the square of any prime. Find m+n.

Note: Without loss of generality, we assume that b > 0. (Otherwise, we can reflect the hexagon across the y axis.) Let the x coordinates of C,D,E, and F be c, d, e, and f , respectively. Note that the y coordinate of C is not 4, since if it were, the fact |AB| = |BC| would imply that A,B, and C are collinear or that c = 0, implying that ABCDEF is concave. Therefore, F = (f, 4). Since $\overrightarrow {A F}=\overrightarrow {C D},C=(c,6)$ and D=(d,10) and so E=(e,8). Because the y coordinates of B,C, and D are 2, 6, and 10, respectively, and |BC| = |CD|, we conclude that b = d. Since $\overrightarrow {A B}=\overrightarrow {E D},e=0.$ Let a denote the side length of the hexagon. Then f < 0. We need to compute

[ABCDEF]=[ABDE]+[AEF]+[BCD]=[ABDE]+2[AEF]

=b.AE+(-f).AE=8(b-f).

Solution:

First Solution: Note that Note that f^2+16=|AF|^2=a^2=|AB|^2=b^2+4. Apply the law of cosines in triangle ABF to obtain 3a^2=|BF|^2=(b-f)^2+4. We have three independent equations in three variables. Hence we can solve this system of equations. The quickest way is to note that

b^2+f^2-2bf+4=(b-f)^2+4=3a^2=a^2+b^2+4+f^2+16,

implying that a^2+16=-2bf.

Squaring both sides gives

a^4+32a^2+16^2=4b^2f^2=4(a^2-4)(a^2-16)=4a^4-80a^2+16^2,

or 3a^4-112a^2=0. Hence $a^2=\frac {112}{3},$ and so $b=\frac {10}{\sqrt {3}}$ and $f=-\frac {8}{\sqrt {3}}.$

Therefore, $[ABCDEF]=8(b-f)=48\sqrt {3},$ and the answer to the problem is 51.

Second Solution: Let $\alpha$ denote the measure (in degrees) of the standard angle formed by the line AB and and the x axis. Then the standard angle formed by the line AF and the x axis is $\beta =120^0+\alpha .$ By considering the y coordinates of B and F, we have $a\sin \alpha=2$ and

$4=a\sin( 120 ^0 +\alpha)=\frac {a\sqrt {3}\cos \alpha}{2}-\frac {a\sin \alpha}{2}=\frac {a\sqrt {3}\cos \alpha}{2}-1,$

by the addition and subtraction formulas.

Hence $a\cos\alpha=\frac {10}{\sqrt {3}}.$ Thus, by considering the x coordinates of B and F, we have

$b=a\cos\alpha=\frac {10}{\sqrt {3}}$ and $f=a\cos(120 ^0 +\alpha)= - \frac {a\cos \alpha}{2} - \frac {a\sqrt {3}\sin \alpha}{2}= - \frac {8}{\sqrt {3}}.$

It follows that $[ABCDEF]=48\sqrt {3}.$

Note: The vertices of the hexagon are

$A=(0,0), B=\Big (\frac {10}{\sqrt {3}},2\Big ), C=\Big (6\sqrt {3}, 6\Big ), D=\Big (\frac {10}{\sqrt {3}},10\Big ), E=(10,8)$

and $F=\Big ( - \frac {8}{\sqrt {3}},4\Big ).$