Advanced Questions on Trignometry Part - 1

by Satish

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Ques: 1 Prove that

1-\cot 23^0=\frac {2}{1-\cot 22^0}

First Solution: We will show that
(1-\cot 3^0)(1-\cot 22^0)=2

Indeed, by the addition and subtraction formulas, we obtain

(1-\cot 23^0)(1-\cot 22^0)=\Big (1-\frac {\cos 23^0}{\sin 23^0}\Big )\Big (1-\frac {\cos 22^0}{\sin 22^0}\Big)

=\frac {\sin 23^0 -cos 23^0}{\sin 23^0}.\frac {\sin 22^0 -cos 22^0}{\sin 22^0}

=\frac {\sqrt {2} \sin (23^0-45^0)\sqrt {2} \sin (22^0-45^0)}{\sin 23^0.\sin 22^0}

=\frac {2\sin (-22^0)\sin (-23^0)}{\sin 23^0.\sin 22^0}
=\frac {2\sin 22^0\sin 23^0}{\sin 23^0\sin 22^0}=2


Second Solution: Note that by the addition and subtraction formula, we have

\frac {\cot 22^0 \cot 23^0-1}{\cot 22^0+\cot 23^0}=\cot (22^0+23^0)=\cot 45^0=1.

Hence \cot 22^0 \cot 23^0-1=\cot 22^0+\cot 23^0, and so

1-\cot 22^0-\cot 23^0+\cot 22^0 \cot 23^0=2

that is, (1-\cot 23^0)(1-\cot 23^0)=2, as desired.



%{font-family:verdana;font-size:17px;color:#150577}*Ques: 2* In triangle ABC, show that

\sin \frac {A}{2}\underline < \frac {a}{b+c}%

Solution: By the extended law of sines, we have

\frac {a}{b+c}=\frac {\sin A}{\sin B+\sin C}

Applying the double-angle formulas and sum-to-product formulas in the above relation gives

\frac {a}{b+c}=\frac {2\sin \frac {A}{2}\cos \frac {A}{2}}{2\sin \frac {B+C}{2}\cos \frac {B-C}{2}}=\frac {\sin \frac {A}{2}}{\cos \frac {B-C}{2}}\underline > \sin \frac {A}{2},

by noting that 0<\cos \frac {B-C}{2}\underline < 1, because 0\underline < |B-C|< 180^0.

Note: By symmetry, we have analogous formulas

\sin \frac {B}{2}\underline < \frac {b}{c+a} and \sin \frac {C}{2}\underline <\frac {c}{a+b}.



%{font-family:verdana;font-size:17px;color:#150577}*Ques: 3* Let f_k(x)=\frac {1}{k}(\sin^k x+\cos^k x) for k=1,2,\cdots Prove that f_4(x)-f_6(x)=\frac {1}{12} for all real numbers x.%

Solution: We need to show that

3(\sin^4 x+\cos^4 x)-2(\sin^6 x+\cos^6 x)=1

for all real numbers x. Indeed, the left-hand side is equal to

3[\sin^2 x+\cos^2 x)^2-2\sin^2 x\cos^2 x]

-2(\sin^2 x+\cos^2 x)(\sin^4 x-\sin^2 x \cos^2 x+\cos^4 x)

=3-6\sin^2 x\cos^2 x-2[(\sin^2 x+\cos^2 x)^2-3\sin^2 x\cos^2 x]=3-2=1.



Ques: 4 A circle of radius 1 is randomly placed in a 15Ã-36 rectangleABCD so that the circle lies completely within the rectangle. Compute the probability that the circle will not touch diagonal AC.

Note :
In order for the circle to lie completely within the rectangle, the center of the circle must lie in a rectangle that is (15-2)x(36-2), or 13×34. The requested probability is equal to the probability that the distance from the circle's center to the diagonalAC is greater than 1, which equals the probability that the distance from a randomly selected point in the 13 × 34 rectangle to each side of triangles ABC and CDA is greater than 1. Let |AB| = 36 and |BC| = 15 (and so |AC| = 39). Drawthree segments that are 1 unit away from each side of triangle ABC and whose endpoints are on the sides. Let E,F, and Gbe the three points of intersection nearest to A,B, and C, respectively, of the three segments. Because the corresponding sides of triangle ABC and EFG are parallel, the two triangles are similar to each other. The desired probability is equal to

\frac {2[EFG]}{13.34}=\Big (\frac {|EF|}{|AB|}\Big )^2.\frac {2[ABC]}{13.34}=\Big (\frac {|EF|}{|AB|}\Big )^2.\frac {15.36}{13.34}=\Big(\frac {|EF|}{|AB|} \Big )^2.\frac {270}{220}.

|<>{border-color:white}. Photo 19569|<>{border-color:white;font-family:verdana}. Because E is equidistant from sides AB and AC, E lies on the bisector of \angle CAB. Similarly, F and G lie on the bisectors of \angle ABC and \angle BCA, respectively. Hence lines AE, BF and CG meet I , the in center of triangle ABC.|

First Solution: Let E_1 and F_1 be the feet of the perpendiculars from E and F to segment AB, respectively. Then |EF|=|E_1F_1|. It is not difficult to see that |BF_1|=|FF_1|=|EE_1|=1. Set \theta =\angle EAB. Then \angle CAB=2\theta,\sin \theta=\frac {5}{12}, \cos 2\theta=\frac {12}{13}, and \tan 2\theta=\frac {5}{12}. By either the double-angle formulas or the half-angle formulas,

\tan 2\theta =\frac {2\tan \theta}{1-\tan^2 \theta} or \tan \theta=\frac {1-\cos 2\theta}{\sin 2 \theta},

and we obtain \tan \theta=\frac {1}{5}. It follows that \frac {|EE_1|}{|AE_1|}=\tan \theta=\frac {1}{5}, or |AE_1|=5. Consequently, |EF|=|E_1F_1|=30. Hence \frac {m}{n}=(\frac {30}{36})^2.\frac {270}{221}=\frac {375}{442}, and m+n=817.

Second Solution: Set A = (0, 0), B = (36, 0), and C = (36, 15). Because E lies on the angle bisector of \angle CAB, \overrightarrow {AE} has the same slope as |\overrightarrow {AB}|\overrightarrow {AC}+|\overrightarrow {AC}|\overrightarrow {AB}=36[36,15]+39[36,0]=[75.36,36.15]=36.15[5,1]; that is, the slope of line AE is \frac {1}{5}. Consequently, |EE_1|=5, and the rest of the solution proceeds like that of the first solution.



Third Solution: Because the corresponding sides of triangles ABC and EFG are parallel, it follows that I is also the incenter of triangle EFG and that the triangles are homothetic (with I as the center). If r is the inradius of triangle ABC, then r âˆ' 1 is the inradius of triangle EFG; that is, the ratio of the similarity between triangles EFG and ABC is \frac {r-1}{r}. Hence the desired probability is (\frac {r-1}{r})^2.\frac {270}{221}.

Note that

r(|AB|+|BC|+|CA|) = 2([AIB]+[BIC]+[CIA]) = 2[ABC] = |AB||.BC|.

Solving the last equation gives r = 6, and so (\frac {5}{6})^2.\frac {270}{221}=\frac {375}{442}.



Ques: 5 Prove that \tan 3a-\tan 2a-\tan a=\tan 3a \tan 2a \tan a for all a\not =\frac {k\pi}{2}, where k is in Z.

Solution: The equality is equivalent to

\tan 3a(1-\tan 2a \tan a)=\tan 2a+\tan a,
or
\tan 3a=\frac {\tan 2a+\tan a}{1-\tan 2a \tan a}.

That is, \tan 3a=\tan(2a+a), which is evident.

Note: More generally, if a_1,a_2,a_3 are real numbers different from \frac {k\pi}{2}, where k is in Z, such that a_1+a_2+a_3=0, then the relation

\tan a_1+\tan a_2+\tan a_3=\tan a_1 \tan a_2 \tan a_3

holds.


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