More Trigonometric Optimization

There is a general rule for solving a system of equations: when you have fewer equations than unknowns, see if you can somehow can get an equation which does the job of several equations. This general principle can, in fact, be applied to the characterization of many other types of triangles. Consider, for example, the following problem:

Question: Suppose ABC is a triangle in which \cos{A}\cos{B}+\sin{A}\sin{B}\sin{C}=1. Prove that

a:b:c=1:1:\sqrt{2} (IITJEE 1986).

Solution: Superficially, the conclusion to be reached involves the sides of the triangle. But, since only their ratios get involved, we can translate it in terms of the angles. We see that a:b:c=1:1:\sqrt{2} is equivalent to saying that

a=b, a^{2}+b^{2}=c^{2}, which, in turn, amounts to saying that the triangle is isosceles and right angled at C. Thus, we have to prove that C=\pi/2 and A=B=\pi/4. (This, by the way, is a helpful habit. See if you can recast the problem, that is, its hypothesis, or its conclusion in various forms. Although mere recasting does not solve the problem, it often inspires you to try certain methods. For instance, now that we have reduced the problem solely in terms of the angles, we can try to solve it by finding the values of \sin{C}, \sin{A}, \cos{A}, etc.)

Now, coming to the problem itself, we are given only one equation (besides the general equation A+B+C=\pi). So somehow we have to cast this equation in a form in which it is equivalent to two separate equations. This can be done by adding and subtracting \sin{A}\sin{B} to get \cos{(A-B)}+\sin{A}\sin{B}(\sin{C}-1)=1 or

(1-\cos{(A-B)})+\sin{A}\sin{B}(1-\sin{C})=0.

The expressions 1-\cos{(A-B)} and 1-\sin{C} are always non-negative. Also, since A, B lie in the interval (0,\pi), \sin{A}, \sin{B} are positive. Therefore the above equation is equivalent to a system of two equations, viz., \cos{(A-B)}=1 and \sin{C}=1. These two equations imply, respectively, that

A=B and C=\pi/2 as desired.

More later,

Nalin Pithwa

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