# Advanced Questions on Trignometry Part - 3

by Satish

Ques: 16 Let ABC be a triangle. Prove that

Solution: By Question 2, we have

The arithmeticâ€"geometric means inequality yields

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 Finally, by (c ) and by the arithmeticâ€" geometric means inequality, we have

implying (d).

Again by Question 2, we have

and analogous formulas for and . Then part (e) follows routinely from the arithmeticâ€"geometric means inequality.

Note: We present another approach to part (a). Note that are all positive. Let It suffices to show that By the arithmeticâ€"geometric means inequality, we have

By Question 15, we have Thus,

Consequently, establishing (a).

Ques: 17 In triangle ABC, show that

Conversely, if x, y, z are positive real numbers such that show that there is an acute triangle ABC such that

Solution: Parts (c.) and (d) follow immediately from (b) because Thus we show only (a) and (b).

(a) Applying the sum-to-product formulas and the fact that we find that

establishing (a).

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

because

Note that It suffices to show that

or which is evident by the sum-to-product formula

From the given equality, we have and thus we may set where Because 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 where Because we know that Because we have implying that Thus, and Therefore, we must have as desired.

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

Using the addition and subtraction formulas, one can easily see that is a nontrivial solution. Hence the determinant of the system is 0; that is,

as desired.