Problem
Let α and β be real numbers such that −π/4 < β < 0 < α < π/4. If sin(α + β) = 1/3 and cos(α − β) = 2/3, find the greatest integer less than or equal to (sinα/cosβ + cosβ/sinα + cosα/sinβ + sinβ/cosα)2.
Solution
To proceed, we’ll try to convert the given expression to a form where it contains only the given functions, i.e. sin(α + β) and cos(α – β).
The first and third terms combine to give (cosα.cosβ + sinα.sinβ)/sinβ.cosβ, which simplifies to 2cos(α − β)/sin2β, or 4/(3sin2β).
Similarly, the second and fourth terms combine to give 4/(3sin2α).
Therefore, the given expression becomes (4/(3sin2β) + 4/(3sin2α))2, or (16/9)(1/sin2α + 1/sin2β)2.
Let’s now consider 1/sin2β + 1/sin2α, which becomes (sin2α + sin2β)/(sin2α.sin2β). Let’s call this A.
The numerator of A equals 2sin(α + β).cos(α − β) or 4/9.
The denominator of A equals (1/2)(cos2(α – β) – cos2(α + β)), which can be written as cos2(α – β) + sin2(α + β) – 1.
Substituting the values, we get the denominator’s value as -4/9. So, A equals -1.
Finally, the given expression or (16/9)A2 equals 16/9, and the greatest integer less than or equal to it equals 1.
In case you need to refer to the various identities we used above, you can refer to the trigonometric identities cheat sheet here.
Comments
A simple, but tedious problem on basic trigonometric identities. Personally, I don’t like forcefully converting the answer to an integer by using the greatest integer function – that’s being lazy! A little more effort could have been taken for the function values so that the given expression comes out to be an integer. Or, maybe ask the value of 9/8 times the given expression (which would’ve evaluated to 2).