Simple trigonometric substitutions with broad results


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1 Simple trigonometric substitutions with broad results Vardan Verdiyan, Daniel Campos Salas Often, the key to solve some intricate algebraic inequality is to simplify it by employing a trigonometric substitution. When we make a clever trigonometric substitution the problem may reduce so much that we can see a direct solution immediately. Besides, trigonometric functions have wellknown properties that may help in solving such inequalities. As a result, many algebraic problems can be solved by using an inspired substitution. We start by introducing the readers to such substitutions. After that we present some wellknown trigonometric identities and inequalities. Finally, we discuss some Olympiad problems and leave others for the reader to solve. Theorem. Let α, β, γ be angles in (0, π). Then α, β, γ are the angles of a triangle if and only if tan α tan β + tan β tan γ + tan γ tan α =. Proof. First of all note that if α = β = γ, then the statement clearly holds. Assume without loss of generality that α β. Because 0 < α + β < π, it follows that there exists an angle in ( π, π), say γ, such that α + β + γ = π. Using the addition formulas and the fact that tan x = cot ( π x), we have tan γ = cot α + β = tan α tan β tan α + tan β, yielding Now suppose that tan α tan β + tan β tan γ + tan γ tan α =. () tan α tan β + tan β tan γ + tan γ tan α =, () for some α, β, γ in (0, π). We will prove that γ = γ, and this will imply that α, β, γ are the angles of a triangle. Subtracting () from () we get tan γ γ = tan. Thus γ γ = kπ for some nonnegative integer k. But γ γ γ + γ < π, so it follows that k = 0. That is γ = γ, as desired. Mathematical Reflections 6 (007)
2 Theorem. Let α, β, γ be angles in (0, π). Then α, β, γ are the angles of a triangle if and only if sin α + sin β + sin γ + sin α sin β sin γ =. Proof. As 0 < α + β < π, there exists an angle in ( π, π), say γ, such that α + β + γ = π. Using the producttosum and the double angle formulas we get sin γ + sin α sin β γ sin = cos α + β ( cos α + β + sin α sin β ) = cos α + β cos α β cos α + cos β = ( ) ( ) sin α + sin β = = sin α sin β. Thus Now suppose that sin α + sin β γ + sin + sin α sin β γ sin =. () sin α + sin β + sin γ + sin α sin β sin γ =, () for some α, β, γ in (0, π). Subtracting () from () we obtain sin γ γ sin + sin α sin β ) γ γ (sin sin = 0, that is ( sin γ ) γ γ γ sin (sin + sin + sin α sin β ) = 0. The second factor can be written as sin γ γ + sin + cos α β cos α + β = sin γ + cos α β, which is clearly greater than 0. It follows that sin γ that α, β, γ are the angles of a triangle. = sin γ, and so γ = γ, showing Mathematical Reflections 6 (007)
3 Substitutions and Transformations T. Let α, β, γ be angles of a triangle. Let A = π α, B = π β, C = π γ. Then A + B + C = π, and 0 A, B, C < π. This transformation allows us to switch from angles of an arbitrary triangle to angles of an acute triangle. Note that cyc(sin α = cos A), cyc(cos α = sin A), cyc(tan α = cot A), cyc(cot α = tan A), where by cyc we denote a cyclic permutation of angles. T. Let x, y, z be positive real numbers. Then there is a triangle with sidelengths a = x + y, b = y + z, c = z + x. This transformation is sometimes called Dual Principle. Clearly, s = x+y+z and (x, y, z) = (s a, s b, s c). This transformation already triangle inequality. S. Let a, b, c be positive real numbers such that ab + bc + ca =. Using the function f : (0, π ) (0, + ), for f(x) = tan x, we can do the following substitution a = tan α, b = tan β, c = tan γ, where α, β, γ are the angles of a triangle ABC. S. Let a, b, c be positive real numbers such that ab + bc + ca =. Applying T to S, we have a = cot A, b = cot B, c = cot C, where A, B, C are the angles of an acute triangle. S3. Let a, b, c be positive real numbers such that a + b + c = abc. Dividing by abc it follows that bc + ca + ab =. Due to S, we can substitute that is where α, β, γ are the angles of a triangle. a = tan α, b = tan β, c = tan γ, a = cot α, b = cot β, c = cot γ, S4. Let a, b, c be positive real numbers such that a + b + c = abc. Applying T to S3, we have a = tan A, b = tan B, c = tan C, Mathematical Reflections 6 (007) 3
4 where A, B, C are the angles of an acute triangle. S5. Let a, b, c be positive real numbers such that a + b + c + abc =. Note that since all the numbers are positive it follows that a, b, c <. Usign the function f : (0, π) (0, ), for f(x) = sin x, and recalling Theorem, we can substitute where α, β, γ are the angles of a triangle. a = sin α, b = sin β, c = sin γ, S6. Let a, b, c be positive real numbers such that a +b +c +abc =. Applying T to S5, we have a = cos A, b = cos B, c = cos C, where A, B, C are the angles of an acute triangle. S7. Let x, y, z be positive real numbers. Applying T to expressions yz (x + y)(x + z), zx (y + z)(y + x), xy (z + x)(z + y), they can be substituted by (s b)(s c), bc (s c)(s a), ca (s a)(s b), ab where a, b, c are the sidelengths of a triangle. Recall the following identities sin α = (s b)(s c) bc Thus our expressions can be substituted by where α, β, γ are the angles of a triangle. sin α, sin β, sin γ,, cos α s(s a) =. bc S8. Analogously to S7, the expressions x(x + y + z) (x + y)(x + z), y(x + y + z) (y + z)(y + x), z(x + y + z) (z + x)(z + y), can be substituted by cos α, cos β, cos γ, where α, β, γ are the angles of a triangle. Mathematical Reflections 6 (007) 4
5 Further we present a list of inequalities and equalities that can be helpful in solving many problems or simplify them. Wellknown inequalities Let α, β, γ be angles of a triangle ABC. Then. cos α + cos β + cos γ sin α + sin β + sin γ 3. sin α + sin β + sin γ cos α + cos β + cos γ cos α cos β cos γ sin α sin β sin γ 8 4. sin α sin β sin γ cos α cos β cos γ cot α + cot β + cot C cos α + cos β + cos γ sin α + sin β + sin C sin α + sin β + sin γ cos α + cos β + cos γ cot α + cot β + cot γ tan α + tan β + tan γ 3 Wellknown identities Let α, β, γ be angles of a triangle ABC. Then. cos α + cos β + cos γ = + 4 sin α sin β sin γ. sin α + sin β + sin γ = 4 cos α cos β cos γ 3. sin α + sin β + sin γ = 4 sin α sin β sin γ 4. sin α + sin β + sin γ = + cos α cos β cos γ For arbitrary angles α, β, γ we have sin α + sin β + sin γ sin(α + β + γ) = 4 sin α + β cos α + cos β + cos γ + cos(α + β + γ) = 4 cos α + β sin β + γ cos β + γ sin γ + α. cos γ + α. Mathematical Reflections 6 (007) 5
6 Applications. Let x, y, z be positive real numbers. Prove that x x + (x + y)(x + z) + y y + (y + z)(y + x) + Solution. The inequality is equivalent to (x + y)(x + z) x = z x + (z + x)(z + y). (Walther Janous, Crux Mathematicorum). (x+y)(x+z) + x Beceause the inequality is homogeneous, we can assume that xy +yz +zx =. Let us apply substitution S: cyc(x = tan α ), where α, β, γ are angles of a triangle. We get ( tan α + tan β ) ( tan α + tan γ ) tan α = sin α, and similar expressions for the other terms. The inequality becomes sin α + sin α + sin β + sin β + sin γ + sin γ, that is + sin α + + sin β + + sin γ. On the other hand, using the wellknown inequality sin α + sin β + sin γ 3 and the CauchySchwarz inequality, we have 9 ( + sin α ) + ( + sin β ) + ( + sin γ ) + sin α, and we are done.. Let x, y, z be real numbers greater than such that x + y + z =. Prove that x + y + z x + y + z. (Iran, 997) Mathematical Reflections 6 (007) 6
7 Solution. Let (x, y, z) = (a +, b +, c + ), with a, b, c positive real numbers. Note that the hypothesis is equivalent to ab + bc + ca + abc =. Then it suffices to prove that a + b + c a + b + c + 3. Squaring both sides of the inequality and canceling some terms yields ab + bc + ca 3. Using substitution S5 we get (ab, bc, ca) = (sin α, sin β, sin γ ), where ABC is an arbitrary triangle. The problem reduces to proving that sin α + sin β + sin γ 3, which is wellknown and can be done using Jensen inequality. 3. Let a, b, c be positive real numbers such that a + b + c =. Prove that ab bc ca c + ab + a + bc + b + ca 3. (Open Olympiad of FML No39, Russia) Solution. The inequality is equivalent to ab ((c + a)(c + b) + bc (a + b)(a + c) + ca (b + c)(b + a) 3. Substitution S7 replaces the three terms in the inequality by sin α, sin β, sin γ. Thus it suffices to prove sin α + sin β + sin γ 3, which clearly holds. 4. Let a, b, c be positive real numbers such that (a + b)(b + c)(c + a) =. Prove that ab + bc + ca 3 4. (Cezar Lupu, Romania, 005) Solution. Observe that the inequality is equivalent to ( ab ) 3 ( 3 4 ) 3 (a + b) (b + c) (c + a). Mathematical Reflections 6 (007) 7
8 Because the inequality is homogeneous, we can assume that ab + bc + ca =. We use substitution S: cyc(a = tan α ), where α, β, γ are the angles of a triangle. Note that (a + b)(b + c)(c + a) = ( ) cos γ cos α cos β = cos α cos β cos γ. Thus it suffices to prove that ( ) or cos α cos β cos γ 4 cos α cos β cos γ 3 3. From the identity 4 cos α cos β cos γ = sin α + sin β + sin γ, the inequality is equivalent to sin α + sin β + sin γ 3 3. But f(x) = sin x is a concave function on (0, π) and the conclusion follows from Jensen s inequality. 5. Let a, b, c be positive real numbers such that a + b + c =. Prove that a + b + c + 3abc., (Poland, 999) ( ) bc Solution. Let cyc x = a. It follows that cyc(a = yz). The inequality becomes x y + y z + x z + 3xyz, where x, y, z are positive real numbers such that xy + yz + zx =. Note that the inequality is equivalent to (xy + yz + zx) + 3xyz + xyz(x + y + z), or 3 x + y + z. Applying substitution S cyc(x = tan α ), it suffices to prove tan α + tan β + tan γ 3. The last inequality clearly holds, as f(x) = tan x is convex function on (0, π), and the conclusion follows from Jensen s inequality. Mathematical Reflections 6 (007) 8
9 6. Let x, y, z be positive real numbers. Prove that x(y + z) + y(z + x) + z(x + y) (x + y)(y + z)(z + x) x + y + z (Darij Grinberg) Solution. Rewrite the inequality as x(x + y + z) (x + y)(x + z) + y(x + y + z) (y + z)(y + x) + Applying substitution S8, it suffices to prove that cos α + cos β + cos γ, z(x + y + z) (z + x)(z + y). where α, β, γ are the angles of a triangle. Using transformation T cyc(a = π α ), where A, B, C are angles of an acute triangle, the inequality is equivalent to sin A + sin B + sin C. There are many ways to prove this fact. We prefer to use Jordan s inequality, that is α π sin α α for all α (0, π ). The conclusion immediately follows. 7. Let a, b, c be positive real numbers such that a + b + c + = 4abc. Prove that a + b + c ab bc ca (Daniel Campos Salas, Mathematical Reflections, 007) Solution. Rewrite the condition as bc + ca + ab + abc = 4. Observe that we can use substitution S5 in the following way ( bc, ca, ) ( = sin α ab, β sin, γ ) sin, Mathematical Reflections 6 (007) 9
10 where α, β, γ are angles of a triangle. It follows that ( a, b, ) ( sin β = sin γ c sin α, sin γ sin α sin β, sin β sin ) α sin γ. Then it suffices to prove that sin β sin γ sin α + sin γ sin α sin β + sin β sin α sin γ 3 sin α + sin β + sin γ. The righthand side of the inequality is well known. For the lefthand side we use trasnformation T backwards. Denote by x = s a, y = s b, z = s c, where s is the semiperimeter of the triangle. The lefthand side is equivalent to x y + z + y x + z + z x + y 3, which a famous Nesbitt s inequality, and we are done. 8. Let a, b, c (0, ) be real numbers such that ab + bc + ca =. Prove that a a + b b + c c 3 ( a + b + ) c. 4 a b c (Calin Popa) Solution. We apply substitution S cyc(a tan A ), where A, B, C are angles of a triangle. Because a, b, c (0, ), it follows that tan A, tan B, tan C (0, ), that is A, B, C are angles of an acute triangle. Note that ( a cyc a = sin A cos ) A = tan A. cos A Thus the inequality is equivalent to tan A + tan B + tan C 3 ( tan A + tan B + ). tan C Now observe that if we apply transformation T and the result in Theorem, we get tan A + tan B + tan C = tan A tan B tan C. Hence our inequality is equivalent to (tan A + tan B + tan C) 3 (tan A tan B + tan B tan C + tan A tan C). This can be written as (tan A tan B) + (tan B tan C) + (tan C tan A) 0, and we are done. Mathematical Reflections 6 (007) 0
11 9. Let x, y, z be positive real numbers. Prove that y + z z + x x + y 6(x + y + z) x y z 3(x + y)(y + z)(z + x). (Vo Quoc Ba Can, Mathematical Reflections, 007) Solution. Note that the inequality is equivalent to (x + y)(z + x) 4(x + y + z) (y + z). x(x + y + z) 3 cyc Let use transfromation T and substitution S8. We get ( ) (x + y)(z + x) cyc (y + z) x(x + y + z) = a cos α = 4R sin α, and 4(x + y + z) 4R(sin α + sin β + sin γ) =, 3 3 where α, β, γ are angles of a triangle with circumradius R. Therefore it suffices to prove that ( 3 sin α + sin β + sin γ ) sin α cos α + sin β cos β + sin γ cos γ. Because f(x) = cos x is a concave function on [0, π], from Jensen s inequality we obtain 3 ( cos α 3 + cos β + cos γ ). Finally, we observe that f(x) = sin x is an increasing function on [0, π], while g(x) = cos x is a decreasing function on [0, π]. Using Chebyschev s inequality, we have ( sin α 3 + sin β + sin γ ) ( cos α + cos β + cos γ ) sin α cos α, and the conclusion follows. Mathematical Reflections 6 (007)
12 Problems for independent study. Let a, b, c be positive real numbers such that a + b + c =. Prove that a b c b + c c + a a + b. (Romanian Mathematical Olympiad, 005). Let a, b, c be positive real numbers such that a + b + c =. Prove that a b + b c + c a 6 (A. Teplinsky, Ukraine, 005) 3. Let a, b, c be positive real numbers such that ab + bc + ca =. Prove that a + b b + c c + a 4. Prove that for all positive real numbers a, b, c, (a + )(b + )(c + ) 9(ab + bc + ca). (Le Trung Kien) (APMO, 004) 5. Let x, y, z be positive real numbers such that x + y + z =. Prove that x + yz + x + yz + x + yz xyz + x + y + z. 6. Let a, b, c be positive real numbers. Prove that b + c + c + a + a + b ( a 4 a b c b + c + b c + a + c ). a + b (APMO, 00) (Mircea Lascu) 7. Let a, b, c be positive real numbers, such that a + b + c = abc. Prove that ab + bc + ca 9(a + b + c). (Belarus, 996) Mathematical Reflections 6 (007)
13 8. Let a, b, c be positive real numbers. Prove that b + c + c + a + a + b abc + a b c (a + b)(b + c)(c + a) (Bui Viet Anh) 9. Let a, b, c be positive real numbers such that a + b + c = abc. Prove that (a )(b )(c ) (Gabriel Dospinescu, Marian Tetiva) 0. Let a, b, c be nonnegative real numbers such that a + b + c + abc = 4. Prove that 0 ab + bc + ca abc. (Titu Andreescu, USAMO, 00) REFERENCES [. ] Titu Andreescu, Zuming Feng, 03 Trigonometry Problems: From the Training of the USA IMO Team, Birkhauser, 004 [. ] Titu Andreescu, Vasile Cirtoaje, Gabriel Dospinescu, Mircea Lascu  Old and New Inequalities, GIL Publishing House, 004 [3. ] Tran Phuong, Diamonds in Mathematical Inequalities, Hanoi Publishing House, 007 [4. ] N.M. Sedrakyan Geometricheskie Neravenstva, Yerevan, 004 [5. ] E. Specht, Collected Inequalities, [6. ] H. Lee, Topics in Inequalities  Theorems and Techniques [7. ] H. Lee, Inequalities through problems [8. ] Crux Mathematicorum and Mathematical Mayhem (Canada) Mathematical Reflections 6 (007) 3
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