# Advanced Questions on Trignometry Part - 1

by Satish

Ques: 1 Prove that

First Solution: We will show that

Indeed, by the addition and subtraction formulas, we obtain

Second Solution: Note that by the addition and subtraction formula, we have

Hence and so

that is, , as desired.

%{font-family:verdana;font-size:17px;color:#150577}*Ques: 2* In triangle ABC, show that

%

Solution: By the extended law of sines, we have

Applying the double-angle formulas and sum-to-product formulas in the above relation gives

by noting that because

Note: By symmetry, we have analogous formulas

and

%{font-family:verdana;font-size:17px;color:#150577}*Ques: 3* Let for Prove that for all real numbers x.%

Solution: We need to show that

for all real numbers x. Indeed, the left-hand side is equal to

Ques: 4 A circle of radius 1 is randomly placed in a 15Ã-36 rectangleABCD so that the circle lies completely within the rectangle. Compute the probability that the circle will not touch diagonal AC.

Note :
In order for the circle to lie completely within the rectangle, the center of the circle must lie in a rectangle that is (15-2)x(36-2), or 13×34. The requested probability is equal to the probability that the distance from the circle's center to the diagonalAC is greater than 1, which equals the probability that the distance from a randomly selected point in the 13 × 34 rectangle to each side of triangles ABC and CDA is greater than 1. Let |AB| = 36 and |BC| = 15 (and so |AC| = 39). Drawthree segments that are 1 unit away from each side of triangle ABC and whose endpoints are on the sides. Let E,F, and Gbe the three points of intersection nearest to A,B, and C, respectively, of the three segments. Because the corresponding sides of triangle ABC and EFG are parallel, the two triangles are similar to each other. The desired probability is equal to

|<>{border-color:white}. |<>{border-color:white;font-family:verdana}. Because E is equidistant from sides AB and AC, E lies on the bisector of Similarly, F and G lie on the bisectors of and respectively. Hence lines AE, BF and CG meet I , the in center of triangle ABC.|

First Solution: Let and be the feet of the perpendiculars from E and F to segment AB, respectively. Then It is not difficult to see that Set Then and By either the double-angle formulas or the half-angle formulas,

or

and we obtain It follows that or Consequently, Hence and

Second Solution: Set and Because E lies on the angle bisector of has the same slope as that is, the slope of line AE is Consequently, and the rest of the solution proceeds like that of the first solution.

Third Solution: Because the corresponding sides of triangles ABC and EFG are parallel, it follows that I is also the incenter of triangle EFG and that the triangles are homothetic (with I as the center). If r is the inradius of triangle ABC, then r âˆ' 1 is the inradius of triangle EFG; that is, the ratio of the similarity between triangles EFG and ABC is Hence the desired probability is

Note that

Solving the last equation gives r = 6, and so

Ques: 5 Prove that for all where k is in Z.

Solution: The equality is equivalent to

or

That is, which is evident.

Note: More generally, if are real numbers different from where k is in Z, such that then the relation

holds.