Advanced Questions on Trignometry Part - 3

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Trigonometry

Advanced Questions on Trignometry Part - 3

Author: Leonardodvin

Ques: 16 Let ABC be a triangle. Prove that

$(a) \quad \sin \frac {A}{2}\sin \frac {B}{2} \sin \frac {C}{2}\underline < \frac {1}{8};$

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

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

$(d) \quad \cos \frac {A}{2}\cos \frac {B}{2}\cos \frac {C}{2}\underline < \frac {3\sqrt {3}}{8};$

$(e) \quad \csc \frac {A}{2}+\csc \frac {A}{2}+ \csc \frac {A}{2}\underline > 6.$

Solution: By Question 2, we have

$\sin \frac {A}{2}\sin \frac {B}{2}\sin \frac {C}{2}\underline < \frac {abc}{(a+b)(b+c)(c+a)}.$

The arithmetic-geometric means inequality yields

$(a+b)(b+c)(c+a)\underline >(2\sqrt {ab})(2\sqrt {bc})(2\sqrt {ca})=8abc.$

Combining the last two equalities gives part (a).

Part (b) then follows from (a) and Question 15. Part (c ) then follows from part (b) by noting that $1-\sin^2 x=\cos^2 x.$ Finally, by (c ) and by the arithmetic- geometric means inequality, we have

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

implying (d).

Again by Question 2, we have

$\csc \frac {A}{2}\underline > \frac {b+c}{a}=\frac {b}{a}+\frac {c}{a}$

and analogous formulas for $\csc \frac {B}{2}$ and $\csc \frac {C}{2}$ . Then part (e) follows routinely from the arithmetic-geometric means inequality.

Note: We present another approach to part (a). Note that $\sin \frac {A}{2}, \sin \frac {B}{2}, \sin \frac {C}{2}$ are all positive. Let $t=\sqrt [3]{\sin \frac {A}{2}\sin \frac {B}{2}\sin \frac {C}{2}}.$ It suffices to show that $t\underline < \frac {1}{2}.$ By the arithmetic-geometric means inequality, we have

$\sin^2 \frac {A}{2}+\sin^2 \frac {B}{2}+\sin^2 \frac {C}{2}\underline > 3t^2.$

By Question 15, we have $3t^2+2t^3\underline <1.$ Thus,

$0\underline >2t^3+3t^2-1=(t+1)(2t^2+t-1)=(t+1)^2(2t-1).$

Consequently, $t\underline < \frac {1}{2},$ establishing (a).

Ques: 17 In triangle ABC, show that

$(a) \quad \sin 2A+\sin 2B+\sin 2C=4\sin A\sin B \sin C;$

$(b) \quad \cos 2A+\cos 2B+\cos 2C=-1-4\cos A\cos B \cos C;$

$(c) \quad \sin^2 A+\sin^2 B+\sin^2 C=2+2\cos A\cos B \cos C;$

$(d) \quad \cos^2 A+\cos^2 B+\cos^2 C+2 \cos A\cos B \cos C=1.$

Conversely, if x, y, z are positive real numbers such that x^2+y^2+z^2+2xyz=1, show that there is an acute triangle ABC such that $x=\cos A, y=\cos B, z=\cos C$

Solution: Parts (c.) and (d) follow immediately from (b) because $\cos 2x=1-2 \sin^2 x=2 \cos^2 x-1.$ Thus we show only (a) and (b).

(a) Applying the sum-to-product formulas and the fact that A+B+C=180^0, we find that

$\sin 2A+\sin 2B+\sin 2C=2 \sin(A+B)\cos(A-B)+\sin 2C$

$=2 \sin C \cos(A-B)+2 \sin C \cos C$

$=2 \sin C[\cos(A-B)-\cos (A+B)]$

$=2 \sin C.[-2 \sin A\ sin(-B)]$

$=4 \sin A\sin B \sin C,$
establishing (a).

(b) By the sum-to-product formulas, we have

$\cos 2A+\cos 2B=2\cos(A+B)\cos(A-B)=-2 \cos C \cos(A-B),$

because

A+B+C=180^0. Note that $\cos 2C+1=2 \cos 2 C.$ It suffices to show that

$-2 \cos C(\cos(A-B)-\cos C)=-4 \cos A\cos B \cos C,$

or $\cos C(\cos(A-B)+\cos(A+B))=2 \cos A\cos B \cos C,$ which is evident by the sum-to-product formula $\cos(A-B)+\cos(A+B)=2 \cos A\ cos B.$

From the given equality, we have $1\underline >x^2,1\underline >y^2,$ and thus we may set $x=\cos A,y=\cos B,$ where $0^0\underline <A,B\underline <90^0.$ Because x^2+y^2+z^2+2xyz is an increasing function of z, there is at most one non-negative value c such that the given equality holds. We know that one solution to this equality is $z=\cos C,$ where C=180^0-A-B. Because $\cos^2 A+\cos^2 B=x^2+y^2\underline <1,$ we know that $\cos^2 B\underline <\sin^2 A.$ Because $0^0<A,B\underline <90^0,$ we have $\cos B \underline <\sin A= \cos (90^0-A),$ implying that $A+B\underline >90^0.$ Thus, $C\underline <90^0$ and $\cos C\underline >0.$ Therefore, we must have $z=\cos C,$ as desired.

Nevertheless, we present a cool proof of part (d). Consider the system of equations

$-x+(\cos B)y+(\cos C)z=0$

$(\cos B)x-y+(\cos A)z=0$

$(\cos C)x+(\cos A)y-z=0.$

Using the addition and subtraction formulas, one can easily see that $(x,y,z)=(\sin A,\sin C,\sin B)$ is a nontrivial solution. Hence the determinant of the system is 0; that is,

$=-1 +2 \cos A\cos B \cos C+\cos^2 A+\cos^2 B+\cos^2 C,$

as desired.