Addition Formulas

Transcript

0:00Introduction

Hello everyone! First, a small puzzle. Tan inverse two and tan inverse three are both positive angles, okay? Add them with the tan addition formula: two plus three, over one minus two times three, gives five over minus five, which is minus one. And tan inverse of minus one is minus pi over four. Two positive angles adding to a negative angle? Something is wrong, right? Today we learn the addition formulas, and the conditions that decide when they work. They come up often in JEE, so let's begin.

0:36Sum of two tan inverses: the formula and its cases

Where does the formula come from? Let capital A be tan inverse x and capital B be tan inverse y; the small minus one up top means inverse, not a power. Then x is tan A and y is tan B, and the tan addition formula gives tan of A plus B equals x plus y, over one minus x y. With me? So we just take tan inverse of both sides, and we are done, right? Let me ask you this: is the sum always tan inverse of that fraction? Not always, and here is why. Capital A and capital B are tan inverse values, so each lies between minus pi over two and pi over two; pi is the Greek letter on screen, and the pointed sign means less than. So their sum can be anywhere between minus pi and pi, but tan inverse only gives answers between minus pi over two and pi over two. Okay? When x y is less than one, the sum stays inside that smaller range, so the plain formula holds: tan inverse x plus tan inverse y equals tan inverse of x plus y, over one minus x y. Right? When both are positive and x y is greater than one, with the sign facing the other way, the sum passes pi over two. So we add pi: the sum is pi plus tan inverse of the fraction. Clear? When both are negative and x y is greater than one, the sum is below minus pi over two, so it is minus pi plus tan inverse of the fraction. Got it? One special case is left: both positive, with x y equal to one. Then the denominator is zero, and the sum is exactly pi over two. With me?

2:19Difference, and the sine and cos inverse sums

Next, the difference. Tan inverse x minus tan inverse y equals tan inverse of x minus y, over one plus x y, for x y greater than minus one. Okay? Now the sine inverse sum. Sine inverse x plus sine inverse y equals sine inverse of the bracket: x times root of one minus y squared, plus y times root of one minus x squared. The tick-shaped sign is the square root. Right? But there is a condition. We need x squared plus y squared less than or equal to one, the less than sign with a line under it, or else x y negative. Otherwise the sum can leave the range of sine inverse. With me? And the cos inverse sum: cos inverse x plus cos inverse y equals cos inverse of x y, minus root of one minus x squared times root of one minus y squared. Note the minus sign, and this one needs x plus y greater than or equal to zero.

3:22Double and triple forms

Clear? Now the double forms. We want two tan inverse x as a single inverse function. Let me ask you this: which substitution makes it easy? Put x equals tan theta, yes, exactly! Theta, the Greek letter, is our angle, so two tan inverse x becomes two theta, and the double angle formulas do the rest. Okay? Two tan inverse x equals sine inverse of two x over one plus x squared, for mod x at most one; the two vertical bars are the modulus, the size of x without its sign. It also equals cos inverse of one minus x squared, over one plus x squared, for x at least zero. Right? And two tan inverse x equals tan inverse of two x over one minus x squared, but only for mod x less than one. Got it? The triple form: three tan inverse x equals tan inverse of three x minus x cubed, over one minus three x squared, for modulus of x below one over root three. Clear? And two sine inverse x equals sine inverse of two x times root of one minus x squared, for modulus of x at most one over root two. With me? Outside these ranges the form changes, and it is not always just plus pi. For example, for x greater than one, two tan inverse x equals pi minus sine inverse of two x over one plus x squared. Find the right form with the same substitution, x equals tan theta or sine theta. Okay?

4:59Example 1: why tan inverse 2 + tan inverse 3 is 3pi/4

Now, the first example. Find tan inverse one half plus tan inverse one third, and then tan inverse two plus tan inverse three. Pause the video and try it first. Done? Okay, let's check. First, x y is one sixth, less than one, so the plain formula works. One half plus one third gives five sixths on top, and one minus one sixth gives five sixths below, so we get tan inverse of one, which is pi over four. Right? Now the second part, our puzzle from the start. Here x y is six, greater than one, and both numbers are positive. Okay? So the answer is tan inverse of five over minus five, that is tan inverse of minus one, which is minus pi over four, right? No, no, that is exactly the trap! Both are positive and x y is greater than one, so we must add pi. Pi plus tan inverse of minus one is pi minus pi over four, which is three pi over four. The answer is positive, and our puzzle is solved. Clear?

6:06Example 2: cos inverse sum, and a sin inverse sum above pi/2

Now a harder one. Find cos inverse of one over root two, plus cos inverse of minus one over root two, using the formula, and check the result directly. Then find sine inverse of four fifths plus sine inverse of twelve thirteenths. Pause the video and try it first. Done? Okay, let's check. Here x plus y is zero, so the cos inverse formula is allowed. Now x y is minus one half, and the two roots multiply to one half. So we get cos inverse of minus one half minus one half, which is cos inverse of minus one, or pi. Okay? Now check it directly. Cos inverse of one over root two is pi over four, and cos inverse of minus one over root two is three pi over four, which add up to pi. The formula agrees, right? Now the second part. Can we put four fifths and twelve thirteenths straight into the sine inverse formula? No, first we must check the condition, always! Here x squared plus y squared is sixteen over twenty five, plus one hundred and forty four over one hundred and sixty nine, and that is more than one, while x y is positive. So the condition fails, and the sum is more than pi over two. With me? The bracket still gives the sine of the sum. The roots are three fifths and five thirteenths, so four fifths times five thirteenths, plus twelve thirteenths times three fifths, gives fifty six over sixty five; the dot means times. Got it? But sine inverse only reaches pi over two, and our sum is bigger. So the sum is the angle between pi over two and pi with this sine, which is pi minus sine inverse of fifty six over sixty five. Always check the condition first.

7:57Quick revision: the condition for each addition formula

Clear? So, quick revision. The tan inverse sum formula works only when x y is less than one; when x y is greater than one, add or subtract pi. The sine inverse sum needs x squared plus y squared at most one, or x y negative; the cos inverse sum needs x plus y at least zero. For double forms, put x equals tan theta or sine theta, and watch the range. Now go and try the practice problems on this page. All the best!