Overview

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. What is sine inverse of sine pi? Sine inverse undoes sine, so the answer is just pi, isn't it? Most students say yes, without a second thought. Hold on to your answer; in a few minutes, you will know the truth. This chapter is the problem bank for the whole trigonometry unit, and that puzzle is one of its concept checks. So let us see how to use this chapter, and then solve two of its problems together. Let's begin!

0:33How to use this chapter

So, let's open this chapter. Part A has sixteen solved problems, at JEE Main level. Okay? Now, how should you use them? The course says: attempt Part A as a thirty minute test, after Chapters one to six; the short dash means to. So finish the first six chapters, and then sit down with a clock. Clear? So, a quick question for you. Thirty minutes for sixteen problems. How much time can you give each problem? Thirty over sixteen is a little less than two. So you get under two minutes a problem, which is good speed practice for the real exam. Right? Then Part B has fourteen longer problems, at JEE Advanced level. Longer problems, so expect more steps in each one. With me? And look at what they mix: identities, equations, inverse functions and triangles, often in the same problem. So you cannot guess the chapter from the question; you have to decide the tool yourself. Got it? And finally, Part C has ten concept checks. These are the statements that are most often tested in true or false form, or as which is correct. Many of them hide a condition, so read each one slowly. Okay?

1:49Part C: check the condition first

Now, two statements from Part C, and the first one is our puzzle. Is sine inverse of, sine x, equal to x for every real x? The small minus one up top means inverse, not one over sine, and the round brackets hold the input. So tell me, is this true or false? The answer is false, and here is the reason. The principal value branch of sine inverse is from minus pi over two to pi over two, so sine inverse gives values only there; pi is the Greek letter, and pi radians is an angle of one hundred and eighty degrees. The square brackets mean both end points are included. Okay? So sine inverse of, sine x, equals x only when x itself lies in that interval. Pi lies outside it. Sine pi is zero, and sine inverse of zero is zero. So sine inverse of sine pi is zero, not pi, and our puzzle is solved. Clear? Now the second statement. Tan inverse x plus tan inverse y equals tan inverse of, x plus y, over one minus x times y, whenever x times y is not equal to one. The slashed equals sign means not equal to. So tell me, is this one true or false? This one is false again, and the reason is a condition. The formula holds only when x times y is less than one; that pointed sign means less than. Right? And when x times y is greater than one, with the sign pointing the other way? If x and y are both positive, add pi; if both are negative, subtract pi. In Part C, the trap is a quick true from habit, so always check the condition. With me?

3:34Example 1: cos 1° + cos 2° + ... + cos 180°

Now, Example one, from Part A, marked easy. Find cos one degree, plus cos two degrees, and so on, up to cos one hundred and eighty degrees. The small circle up top means degrees, and the three dots stand for all the terms in between. Remember, under two minutes. Pause the video and try it first. Done? Okay, let's check. Here is the key fact: cos of, one hundred and eighty degrees minus k degrees, equals minus cos k degrees, where k is just a number. Cosine is negative in the second quadrant, right? So pair each angle with its partner. Cos k degrees plus cos of, one hundred and eighty minus k, degrees, is zero, for k from one up to eighty nine; the dots again mean every value in between. That is eighty nine pairs, which covers one hundred and seventy eight terms. Okay? So everything cancels, and the answer is zero, right? No, no, careful, because that is exactly the trap here! There are one hundred and eighty terms, but only one hundred and seventy eight are paired. Two are left over: cos ninety degrees, which is zero, and cos one hundred and eighty degrees, which is minus one. Clear? So the sum is zero from the pairs, zero from cos ninety degrees, and minus one from cos one hundred and eighty degrees. The total is minus one. Always count the terms that have no partner. Got it?

5:03Example 2: sin x + sin 2x + sin 3x = cos x + cos 2x + cos 3x

Now a harder one, from Part B, marked hard. Solve sine x plus sine two x plus sine three x, equals cos x plus cos two x plus cos three x, for x from zero to two pi, both ends included. Pause the video and try it first. Done? Okay, let's check. There are three sines on the left. Which two would you pair up, and why? Yes, exactly, the outer two, x and three x, because their average is two x! Sine x plus sine three x is two times sine two x, times cos x. And cos x plus cos three x is two times cos two x, times cos x. Okay? Now add the middle terms back. The left side is sine two x times, two cos x plus one. The right side is cos two x times, two cos x plus one. The same bracket on both sides! Clear? So we cancel that bracket from both sides, and we just get sine two x equals cos two x, right? Oh no, careful, because that bracket can be zero, and cancelling it throws those solutions away! Bring everything to one side and factor instead: two cos x plus one, times, sine two x minus cos two x, equals zero. Got it? Take the first factor equal to zero, so cos x equals minus one half. From zero to two pi, that gives x equals two pi over three, and four pi over three. These are exactly the two solutions the cancelling would have lost. With me? Now take the second factor equal to zero, so sine two x equals cos two x. Cos two x is not zero here, because sine and cos of the same angle are never both zero, so we may divide, and tan two x equals one. So two x equals n pi plus pi over four, where n is any integer. Right? Now x runs from zero to two pi, so two x runs from zero to four pi, and n can be zero, one, two or three only. Dividing by two, x is pi over eight, five pi over eight, nine pi over eight, and thirteen pi over eight. These four, plus the two from cos x, give six solutions in all, not four. Okay?

7:23Quick revision: check conditions, never cancel a zero factor

So, quick revision. Part A is a thirty minute test, Part B is at JEE Advanced level, and Part C tests statements. Always check the condition: sine inverse of, sine x, equals x only for x from minus pi over two to pi over two. And count the unpaired terms, and never cancel a factor that can be zero. Now open Part A, set a thirty minute clock, and start solving. All the best!