Systems, Bounds and Inequalities

Transcript

0:00Introduction

Hello everyone! Here is a really scary equation. Sine to the power four of x equals one, plus tan to the power eight of four x. Two different angles, a fourth power and an eighth power. Which identity would you use? Honestly, none of them. It gets solved in two lines, just by comparing sizes. Today we learn bounds, systems and inequalities. They are useful in JEE problems, so let's begin.

0:30Bounds: squeeze the two sides

So, the first idea is bounds. Suppose the left side can never be more than some number, capital M, and the right side can never be less than it. The sign with a line under it means less than or equal to, and the other one means greater than or equal to. Now tell me, when can the two sides be equal? Only when both are exactly M, yes, exactly! One side cannot go up, the other cannot come down, so they meet only there. One hard equation breaks into two easy ones. Okay? Now, the bounds we use most. The simplest is sine x at most one, for every x. Right? Next, small a cos x plus small b sine x is at most the square root of a squared plus b squared. The tick-shaped sign is the square root, and the small two up top means squared. We met this in the standard methods, remember? And the third bound says t plus one over t is at least two. But wait, is this true for every t? Try t equal to minus one. With t equal to minus one, we get minus one plus minus one, which is minus two, less than two! So this bound needs t positive; the sign without the line means strictly greater than. And the two sides are equal only when t is one. Clear?

1:55Systems: x+y and x-y as the unknowns

Alright, the second idea is systems in x and y. Often one equation gives x plus y, so we make x plus y and x minus y the unknowns, using sum-to-product formulas. For example, cos x plus cos y equals two cos of x plus y over two, times cos of x minus y over two. Okay? And for a product, two sine x sine y equals cos of x minus y, minus cos of x plus y. Both bring in only x plus y and x minus y. With me? Now, if the first equation already gives x plus y, what is left to find? Only x minus y, yes! Put the known x plus y into the formula, solve for x minus y, then add and subtract to get x and y separately. Got it? But be careful here. Sometimes the formula asks for a cosine bigger than one. Cosine never goes above one, so then there is no solution at all. Right?

2:55Inequalities: equation first, then the graph

Now the third idea, inequalities, where we always solve the equation first. For sine x greater than one half, first find where sine x equals one half: from zero to two pi, at pi over six and at five pi over six. Pi radians is one hundred and eighty degrees, and the square brackets mean both those ends are included. Okay? Now picture the sine curve from zero to two pi. Between which two angles is it above one half? Between pi over six and five pi over six, yes, exactly! The round brackets mean the ends are left out, because there sine x is exactly one half, not greater. Clear? Sine repeats every two pi, so add two n pi to both ends. Here n is any integer; the curly e-shaped sign means belongs to, and the double-struck Z is the set of integers. With me? For tan, the period is pi, so we add n pi instead. In one period, tan x is greater than one from pi over four up to pi over two. Pi over two stays out, because tan is not defined there. Got it?

4:04Example 1: sin x to the 4th = 1 + tan 4x to the 8th

Now, let's finish our scary equation from the start. Solve sine to the power four of x equals one, plus tan to the power eight of four x. The small four and the small eight up top are powers. Pause the video and try it first. Done? Okay, let's check. Sine x lies between minus one and one, so its fourth power is at most one. And tan to the power eight is never negative, so the right side is at least one. Right? So the left side is at most one, and the right side is at least one. What must both sides equal? Both must equal one, yes, exactly! So sine to the power four of x equals one, and tan four x equals zero. Clear? Sine to the power four of x is one, so sine x is plus or minus one; that is the plus sign over the minus sign, and the double arrow means implies. That happens at the odd multiples of pi over two, so x equals two n plus one, times pi over two. Okay? But both conditions must hold, so check the second one too. Four x is then four n plus two, times pi, a whole multiple of pi. So tan four x is zero, and it is defined there, since cos four x is one. Right? So the answer is x equals two n plus one, times pi over two, with n any integer. Two lines, just as promised. With me?

5:33Example 2: sin x + root 3 cos x at least 1 in 0 to 2pi

Now a harder one, JEE level. Find all x from zero to two pi, with sine x plus root three cos x at least one. Root three is the square root of three. Hint: write the left side as one sine first. Pause the video and try it. Done? Okay, let's check. Take out a factor of two: then we need one half sine x plus root three over two cos x. One half is cos of pi over three, and root three over two is sine of pi over three. So sine x plus root three cos x is two sine of x plus pi over three, and we need sine of x plus pi over three at least one half. Okay? Now, sine is at least one half from pi over six to five pi over six. So x plus pi over three lies between those, and x runs from minus pi over six to pi over two. Inside our range, that is zero to pi over two. So we are done, right? No, no, wait, this is the trap! Since x runs from zero to two pi, x plus pi over three runs from pi over three to seven pi over three. That is more than one full turn, so sine comes back above one half. With me? So in this range, the sine is at least one half from pi over three to five pi over six, and again from thirteen pi over six to seven pi over three. Thirteen pi over six is just two pi plus pi over six. The cup-shaped sign means union, so either piece works. Clear? Now subtract pi over three from every end. Five pi over six minus pi over three gives pi over two, and thirteen pi over six minus pi over three gives eleven pi over six. So x lies from zero to pi over two, or from eleven pi over six to two pi. That second piece is what the trap missed. Got it?

7:27Quick revision: bounds, systems and inequalities

So, quick revision. For bounds, if one side is at most M and the other is at least M, both equal that number. For systems, use sum-to-product to find x plus y and x minus y. For inequalities, solve the equation, read one period, then add two n pi, or n pi. Now go and try the practice problems on this page. All the best!