Greatest and Least Values

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take three sine theta, minus four cos theta. What is the largest value it can take? Sine theta is at most one, and cos theta is at least minus one, so three plus four gives seven, right? Hmm, I would not be so sure about that. In this lesson we will find the true answer, and learn to find greatest and least values in trigonometry. Such questions come up often in JEE, so let's begin.

0:31The trap: is the greatest value 7?

So, here is our puzzle. Theta, the Greek letter, is the angle, and it can be anything. The quick answer says three plus four, which is seven. Okay? Now look at what seven needs. Sine theta must be one, and cos theta must be minus one, both at the same time. Let me ask you this: is that possible? No, it is never possible! Sine squared theta plus cos squared theta is always one; the small two after sine and cos means squared. So when sine theta is one, cos theta must be zero. The two terms can never be at their best together, so seven is never reached. Clear?

1:12The key result: a cos theta + b sin theta

So here is the key result. Small a and small b are fixed numbers. Small a cos theta plus small b sine theta equals the root of a squared plus b squared, times cos of theta minus phi. The tick-shaped sign is the square root, the round brackets hold the angle, and phi is another Greek letter. With me? And phi is a fixed angle: cos phi is small a over that root, and sine phi is small b over that root, so tan phi is small b over small a. Use both, not just the tan, to get phi in the right quadrant. So the whole expression is a single cos, stretched by that root. Okay? Now tell me: the cos of any angle lies between which two numbers? Between minus one and one, yes, exactly! Multiply through by that root. So the expression lies between minus that root and plus that root; the sign with a line under it means less than or equal to. Write this down. Got it?

2:12Careful: when the angle is restricted

But careful, this result has a trap. The greatest value needs cos of theta minus phi to be exactly one. So the bound is attained only if the angle that gives it is allowed. Right? So when theta is restricted to an interval, do not quote plus or minus the root straight away. First find where theta minus phi lies, and read the greatest and least values from there. Example two shows exactly this.

2:39Squares, products and quadratics

Clear? Now, suppose you get small a sine squared theta, plus small b sine theta cos theta, plus small c cos squared theta. The trick is to change everything to two theta. Okay? Use the double angle forms. Sine squared theta is one minus cos two theta, over two. Cos squared theta is one plus cos two theta, over two. And sine theta cos theta is sine two theta, over two. With me? Put these in, and you get a constant, a cos two theta term and a sine two theta term. That is our key result again, with two theta in place of theta. Nice, right? Next, a quadratic in sine theta. Put t equal to sine theta, and complete the square. Now let me ask: can t take any real value here? No, t is a sine, so it lies between minus one and one; the symbol like a rounded E means belongs to, and the square brackets include both ends. So the greatest or least value may come at an end, where t is minus one or t is one. Got it?

3:43Topper's tools: positive terms and triangles

Now two tools that toppers use. When all the terms are positive, use the AMGM inequality: the arithmetic mean is at least the geometric mean. So x plus y is at least two root of x times y; this sign means greater than or equal to. Okay? And a sharper form: p squared over x, plus q squared over y, is at least p plus q, whole squared, over x plus y, for positive x and y. It is perfect for sums like four over s, plus one over one minus s. Right? In any triangle with angles capital A, B and C, these bounds hold. Sine A sine B sine C is at most three root three over eight, and cos A cos B cos C is at most one over eight. With me? Also, cos A plus cos B plus cos C is at most three over two, and sine A over two, times sine B over two, times sine C over two, the sines of the half angles, is at most one over eight. Now guess: which triangle makes all four equalities? The equilateral triangle, yes, exactly! All three angles equal, sixty degrees each. Clear?

5:00Example 1: range of 3 sin theta - 4 cos theta + 7

Now, remember our puzzle? Let's finish it. Find the range of three sine theta, minus four cos theta, plus seven. Pause the video and try it first. Done? Okay, let's check. Match it with small a cos theta plus small b sine theta. Here small a is minus four, and small b is three. So the root of sixteen plus nine is the root of twenty five, which is five. Okay? So three sine theta minus four cos theta lies between minus five and five. Puzzle solved: five, not seven, because sine and cos never reach their extremes together. Right? Now add seven everywhere. Minus five plus seven is two, and five plus seven is twelve. With me? So the range is the closed interval from two to twelve. Theta can be any angle here, so both ends are reached, and that is it!

5:57Example 2: max and min of sin x + root 3 cos x on [0, pi/2]

Now a harder one, JEE level. Find the greatest and least values of sine x plus root three cos x, for x from zero to pi over two, both ends included; pi is the Greek letter on screen. Pause the video and try it first, and here is a hint: remember the trap. Done? Okay, let's check. The root of one plus three is two. So the greatest value is two and the least is minus two, right? Not so fast, that is the trap! Here x is restricted, so check before you quote these. So first, write it as one sine. Take two outside; inside, we get sine x times one half, plus cos x times root three over two; the dot means times. Okay? Now one half is cos pi over three, and root three over two is sine pi over three. So, by the formula for the sine of a sum, the bracket is sine of x plus pi over three, and the whole thing is two sine of x plus pi over three. With me? Now, x runs from zero to pi over two. So where does x plus pi over three lie? Just add pi over three to both ends. It runs from pi over three to five pi over six; the double arrow means implies. Right? On this interval, sine starts at root three over two, rises to one at pi over two, then falls to one half at five pi over six. So the smallest sine here is one half, not minus one! Clear? So the greatest value is two times one, which is two, at x equal to pi over six. And the least is two times one half, which is one, at x equal to pi over two. Two and one, so the trap is avoided!

7:51Quick revision: root(a^2+b^2) bound, restricted angle, AM-GM

So, quick revision. Small a cos theta plus small b sine theta lies between minus and plus the root of a squared plus b squared. If the angle is restricted, first find where the combined angle lies. Squares go to two theta, quadratics keep t between minus one and one, and for positive terms, try AMGM first. Now go and try the practice problems on this page. All the best!