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Leonardodvin

Advanced Questions on Trignometry Part - 4

Author: Leonardodvin

Ques: 18 In triangle ABC, show that

$(a) \quad 4R=\frac {abc}{[ABC]};$

$(b) \quad 2R^2 \sin A\sin B \sin C=[ABC];$

$(c) \quad 2 R \sin A\sin B \sin C=r(\sin A+\sin B+\sin C);$

$(d) \quad r=4R \sin\frac {A}{2}\sin \frac {B}{2} \sin \frac {C}{2};$

$(e) \quad a\cos A+b \cos B+c\cos C=\frac {abc}{2R^2}.$

Solution: By the extended law of sines,

$R=\frac {a}{2 \sin A}=\frac {abc}{2bc\sin A}=\frac {abc}{4[ABC]},$

establishing (a).

By the same token, we have

$2R^2 \sin A\sin B \sin C=\frac {1}{2}.(2R\sin A)(2R\sin B)(\sin C)=\frac {1}{2}ab \sin C=[ABC],$

which is (b).

Note that: $2[ABC]=bc \sin A=(a+b+c)r.$

By the extended law of sines, we obtain

$4R^2 \sin A\sin B \sin C=bc \sin A=r(a+b+c)=2rR(\sin A+\sin B+\sin C),$

from which (c ) follows.

By the law of cosines,

$\cos A=\frac {b^2+c^2-a^2}{2bc}.$

Hence, by the half-angle formulas, we have

$\sin^2 \frac {A}{2}=\frac {1-\cos A}{2}=\frac {1}{2}-\frac {b^2+c^2-a^2}{4bc}=\frac {a^2-(b^2+c^2-2bc)}{4bc}$

$=\frac {a^2-(b-c)^2}{4bc}=\frac {(a-b+c)(a+b-c)}{4bc}=\frac {(2s-2b)(2s-2c)}{4bc}=\frac {(s-b)(s-c)}{bc},$

where 2s = a + b + c is the perimeter of triangle ABC. It follows that

$\sin \frac {A}{2}=\sqrt {\frac {(s-b)(s-c)}{bc}},$

and the analogous formulas for $\sin \frac {B}{2}$ and $\sin \frac {C}{2}.$ Hence

$\sin \frac {A}{2}\sin \frac {B}{2}\sin \frac {C}{2}=\frac {(s-a)(s-b)(s-c)}{abc}=\frac {s(s-a)(s-b)(s-c)}{sabc}=\frac {[ABC]^2}{sabc}$

by Heron's formula. It follows that

$\sin \frac {A}{2}\sin \frac {B}{2}\sin \frac {C}{2}=\frac {[ABC]}{s}.\frac {[ABC]}{abc}=r.\frac {1}{4R},$

from which (d) follows.

Now we prove (e). By the extended law of sines, we have a

$\cos A=2R \sin A.\cos A=R\sin 2A.$

Likewise, $b \cos B=R\sin 2 B$ and $c \cos C=R \sin 2 C.$

By (a) and (b), we have

$4R \sin A\sin B \sin C=\frac {abc}{2R^2}.$

It suffices to show that

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

which is Question 17(a).

Ques: 19 Let s be the semiperimeter of triangle ABC. Prove that

$(a) \quad s=4R\cos \frac {A}{2}\cos \frac {B}{2}\cos \frac {C}{2};$

$(b) \quad s\underline < \frac {3\sqrt {3}}{2}R.$

Solution: It is well known that rs=[ABC], or $s=\frac {[ABC]}{r}.$

By Question 18 (b) and (d), part (a) follows from

$s=\frac {R\sin A\sin B\sin C}{2\sin \frac {A}{2}\sin \frac {B}{2}\sin \frac {C}{2}}=4R\cos \frac {A}{2}\cos \frac {B}{2}\cos \frac {C}{2}$

by the double-angle formulas.

We conclude part (b) from (a) and Question 16 (d).

Ques: 20 In triangle ABC, show that

$(a) \quad \cos A+\cos B+\cos C=1+4\sin \frac {A}{2}\sin \frac {B}{2} \sin \frac {C}{2};$

$(b) \quad \cos A+\cos B+\cos C\underline < \frac {3}{2}.$

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

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

and

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

It suffices to show that

$2\sin \frac {C}{2}\Big [\cos \frac {A-B}{2}- \cos \frac {A+B}{2}\Big ]=4\sin \frac {A}{2}\sin \frac {B}{2} \sin \frac {C}{2},$

or,

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

which follows from the sum-to-product formulas, and hence (a) is established. Recalling Question 18 (c.), we have

$\cos A+\cos B+\cos C=1+\frac {r}{R}. \qquad \qquad \qquad \qquad (*)$

Euler's formula states that |OI|^2=R^2-2Rr, where O and I are the circumcenter and incenter of triangle ABC. Because $|OI|^2\underline >0,$ we have $R\underline >2r,$ or $\frac {r}{R}\underline < \frac {1}{2},$ from which (b) follows.

Note: Relation (âˆ-) also has a geometric interpretation.

As shown in the Figure, let O be the circumcenter, and let A_1,B_1,C_1

be the feet of the perpendiculars from O to sides BC,CA,AB, respectively. (Thus A_1,B_1,C_1

are the midpoints of sides BC,CA,AB, respectively.) Because $\angle AOB=2C$ and triangle AOB is isosceles with |OA|=|OB|=R,

we have $|OC_1|=R\cos C.$ Likewise, $|OB_1|=R\cos B$ and $|OA_1|=R\cos A.$

It suffices to show that

|OA_1|+|OB_1|+|OC_1|=R+r.

Note that

|OA|=|OB|=|OC|=R and |BA_1|=|A_1C|,|CB_1|=|B_1A|,|AC_1|=|C_1B|.

Hence |AB|=2|A_1B_1|,|BC|=2|B_1C_1|,|CA|=2|C_1A_1|.

Let s denote the semiperimeter of triangleABC.Applying Ptolemy's theorem to cyclic quadrilaterals OA_1CB_1,OB_1AC_1,OC_1BA_1 yields

|A_1B_1|.|OC|=|A_1C|.|OB_1|+|CB_1|.|OA_1|,

|B_1C_1|.|OA|=|B_1A|.|OC_1|+|AC_1|.|OB_1|,

|C_1A_1|.|OB|=|C_1B|.|OA_1|+|BA_1|.|OC_1|.

Adding the above gives

Rs=|OA_1|(s-|A_1B|)+|OB_1|(s-|B_1C|)+|OC_1|(s-|C_1A|)

=s(|OA_1|+|OB_1|+|OC_1|)-[ABC]

=s(|OA_1|+|OB_1|+|OC_1|)-rs,

from which our desired result follows.