Advanced Questions on Trignometry Part - 2

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Advanced Questions on Trignometry Part - 2

Author: Leonardodvin

Ques: 6 Let a, b, c, d be numbers in the interval $[0,\pi]$ such that

$\sin a+7 \sin b=4(\sin c+2 \sin d),$

$\cos a+7 \cos b=4(\cos c+2 \cos d).$

Prove that $2 \cos(a-d)=7 \cos(b-c).$

Solution: Rewrite the two given equalities as

$\sin a-8 \sin d=4 \sin c-7 \sin b,$

$\cos a-8 \cos d=4 \cos c-7 \cos b.$

By squaring the last two equalities and adding them, we obtain $1+64-16(\cos a \cos d+\sin a \sin d)=16+49-56(\cos b \cos c+\sin b \sin c),$ and the conclusion follows from the addition formulas.

Ques: 7 Express $\sin(x-y)+\sin(y-z)+\sin(z-x)$ as a monomial.

Solution: By the sum-to-product formulas, we have

$\sin(x-y)+\sin(y-z)=2 \sin \frac {x-z}{2}\cos \frac {x+z-2y}{2}.$

By the double-angle formulas, we have

$\sin(z-x)=2 \sin \frac {z-x}{2}\cos \frac {z-x}{2}.$

Thus

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

$=2\sin \frac {x-z}{2}\Big [\cos \frac {x+z-2y}{2}-\cos \frac {z-x}{2}\Big ]$

$=-4\sin \frac {x-z}{2}\sin \frac {z-y}{2}\sin \frac {x-y}{2}$

$=-4\sin \frac {x-y}{2}\sin \frac {y-z}{2}\sin \frac {z-x}{2}$

by the sum-to-product formulas.

Note: In exactly the sameway, we can showthat if a, b, and c are real numbers with a + b + c = 0, then

$\sin a+\sin b+\sin c=-4\sin \frac {a}{2}\sin \frac {b}{2}\sin \frac {c}{2}.$

In this question, we have a=x-y,b=y-z, and c=z-x.

Ques: 8 Prove that $(4\cos^2 9^0-3)(4\cos^2 27^0-3)=\tan 9^0.$

Solution: We have $\cos 3x=4\cos^3x-3\cos x,$ so $4\cos^2 x-3=\frac {\cos 3x}{\cos x}$ for all $x\not = (2k+1).90^0, k\in Z.$ Thus

$(4\cos^29^0-3)(4\cos^227^0-3)=\frac {\cos 27^0}{\cos 9^0}.\frac {\cos 81^0}{\cos 27^0}=\frac {\cos 81^0}{\cos 9^0}=\frac {\sin 9^0}{\cos 9^0}=\tan 9^0$

as desired.

Ques: 9 Prove that $\Big (1+\frac {a}{\sin x}\Big ) \Big (1+\frac {b}{\cos x}\Big ) \underline > \Big (1+\sqrt {2ab}\Big )^2$
for all real numbers a, b, x with $a, b\underline >0$ and $0<x<\frac {\pi}{2}.$

Solution: Expanding both sides, the desired inequality becomes

$1+\frac {a}{\sin x}+\frac {b}{\cos x}+\frac {ab}{\sin x \cos x}>1+2ab+2\sqrt {2ab}.$

By the arithmetic-geometric means inequality, we obtain

$\frac {a}{\sin x}+\frac {b}{\cos x}\underline > \frac {2\sqrt {ab}}{\sqrt {\sin x \cos x}}.$

By the double-angle formulas, we have $\sin x \cos x =\frac {1}{2} \sin 2x \underline < \frac {1}{2},$ and so

$\frac {2\sqrt {ab}}{\sqrt {\sin x \cos x}}\underline >2\sqrt {2ab}$

and

$\frac {ab}{\sin x \cos x}\underline > 2ab.$

Combining the last three inequalities gives the the desired result.

Ques: 10 In triangle ABC, $\sin A+\sin B+\sin C\underline <1.$ Prove that

$min \{A+B, B+C, C+A\}<30^0.$

Solution: Without loss of generality, we assume that $A\underline >B\underline >C.$ We need to prove that B+C<30^0. The law of sines and the triangle inequality (b+c>a) imply that $\sin B+\sin C>\sin A,$ so $\sin A+\sin B+ \sin C>2 \sin A.$ It follows that $\sin A<\frac {1}{2},$ and the inequality $A\underline >\frac {A+B+C}{3}=60^0$ gives that A>150^0; that is, B+C<30^0, as desired.

Ques: 11 Let ABC be a triangle. Prove that

$\tan \frac {A}{2}\tan \frac {B}{2}+\tan \frac {B}{2}\tan \frac {C}{2}+\tan \frac {C}{2}\tan \frac {A}{2}=1$

Solution: By the addition and subtraction formulas, we have

$\tan \frac {A}{2}+\tan \frac {B}{2}=\tan \frac {A+B}{2}\Big (1-\tan \frac {A}{2}\tan \frac {B}{2}\Big ).$

Because $A+B+C=180^0, \frac {A+B}{2}=90^0-\frac {C}{2},$ and so $\tan \frac {A+B}{2}=\cot \frac {C}{2}.$

Thus

$\tan \frac {A}{2}\tan \frac {B}{2}+\tan \frac {B}{2}\tan \frac {C}{2}+\tan \frac {C}{2}\tan \frac {A}{2}$

$=\tan \frac {A}{2}\tan \frac {B}{2}+\tan \frac {C}{2}\cot \frac {C}{2}\Big (1-\tan \frac {A}{2}\tan \frac {B}{2}\Big )$

$=\tan \frac {A}{2}\tan \frac {B}{2}+1-\tan \frac {A}{2}\tan \frac {B}{2}=1$

as desired.

Note: An equivalent form of this equation is:

$\cot \frac {A}{2}+\cot \frac {B}{2}+\cot \frac {C}{2}=\cot \frac {A}{2}\cot \frac {B}{2}\cot \frac {C}{2}.$

Ques: 12 Let ABC be a triangle. Prove that

$\tan \frac {A}{2}\tan \frac {B}{2}\tan \frac {C}{2}\underline < \frac {\sqrt {3}}{9}.$

Solution: By the arithmetic-geometric means inequality, we have

$1=\tan \frac {A}{2}\tan \frac {B}{2}+\tan \frac {B}{2}\tan \frac {C}{2}+\tan \frac {C}{2}\tan \frac {A}{2}$

$\underline >3 \sqrt [3]{\Big(\tan \frac {A}{2}\tan \frac {B}{2}\tan \frac {C}{2} \Big )^2},$

from which follows the desired equation.

Ques: 13 Let ABC be an acute-angled triangle. Prove that

$\tan A+\tan B +\tan C=\tan A \tan B \tan C;$
$\tan A \tan B \tan C \underline >3\sqrt {3}.$

Solution: Note that because of the condition $A,B,C \not =90^0,$ all the above expressions are well defined.

The proof of the identity in part (1) is similar to that of Question 11. By the arithmetic-geometric means inequality,

$\tan A+\tan B+\tan C\underline >3\sqrt [3]{\tan A \tan B \tan C}.$

By (1), we have

$\tan A \tan B \tan C \underline > \sqrt [3]{\tan A \tan B \tan C},$

from which follows equation (2)

Note: Indeed, the identity in (1) holds for all angles A,B,C with $A+B+C=m\pi$ and $A,B,C\not = \frac {k\pi}{2},$ where k and m are in Z.

Ques: 14 Let ABC be a triangle. Prove that

$\cot A\cot B+\cot B \cot C+\cot C \cot A=1.$

Conversely, prove that if x, y, z are real numbers with xy+yz+zx=1, then there exists a triangle ABC such that $\cot A=x, \cot B=y,$ and $\cot C=z.$

Solution: If ABC is a right triangle, then without loss of generality, assume that A=90^0. Then $\cot A=0$ and B+C=90^0, and so $\cot B \cot C=1,$ implying the desired result.

If $A,B,C\not =90^0,$ then $\tan A\ tan B \tan C$ is well defined. Multiplying both sides of the desired identity by $\tan A\tan B \tan C$ reduces the desired result to Question 13(1).

The second claim is true because $\cot x$ is a bijective function from the interval (0^0,180^0) to $(-\infty ,\infty).$

Ques: 15 Let ABC be a triangle. Prove that

$\sin^2 \frac {A}{2}+\sin^2 \frac {B}{2}+\sin^2 \frac {C}{2}+2 \sin \frac {A}{2}\sin \frac {B}{2}\sin \frac {C}{2}=1.$

Conversely, prove that if x, y, z are positive real numbers such that

x^2+y^2+z^2+2xyz=1,

then there is a triangle ABC such that $x=\sin \frac {A}{2},y=\sin \frac {B}{2},$ and $z=\sin \frac {C}{2}.$

Solution: Solving the second given equation as a quadratic in x gives

$x=\frac {-2yz+\sqrt {4y^2z^2-4(y^2+z^2-1)}}{2}=-yz+\sqrt {(1-y^2)(1-z^2)}.$

We make the trigonometric substitution $y=\sin \mu$ and $z=\sin \upsilon,$ where $0^0< \mu, \upsilon < 90^0.$ Then

$x=-\sin \mu \sin \upsilon +\cos \mu \cos \upsilon=\cos (\mu + \upsilon).$

Set $\mu= \frac {B}{2}, \upsilon=\frac {C}{2}$ and A=180^0-B-C.

Because

$1\underline > y^2+z^2=\sin^2\frac {B}{2}+\sin^2\frac {C}{2},\cos^2\frac {B}{2}\underline > \sin^2 \frac {C}{2}.$

Because

$0^0 < \frac {B}{2}, \frac {C}{2}< 90^0, \cos \frac {B}{2} > \sin \frac {C}{2}=\cos (90^0-\frac {C}{2}),$ implying that $\frac {B}{2} < 90^0-\frac {C}{2},$ or B+C < 180^0. Then $x=\cos (\mu +\upsilon)=\sin \frac {A}{2}, y=\sin \frac {B}{2},$ and $z=\sin \frac {C}{2},$ where A, B and C are the angles of a triangle.

If ABC is a triangle, all the above steps can be reversed to obtain the first given identity.