Mixed Problems: Set 1 (Values, Identities, Bounds)

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Three sine theta plus four cos theta can never go above five. And five times sine of theta plus pi over two can never go above five either. So the greatest value of their sum must be ten, right? Hmm, keep that guess ready, because Example two settles it. This is Set one of our mixed trigonometry problems, in the JEE Main pattern: values, identities and bounds. Questions like these come up often, so let's begin.

0:34Set 1: what it holds

So, here is what Set one holds. Fifteen solved problems on values, identities and bounds. They are all mixed up, so you choose the tool yourself, as in the exam. Okay? And the course target is one to two minutes per problem. So first, let us sharpen three tools from the earlier chapters, and then we solve two problems together. With me?

0:57Tool 1: given tan θ, divide by cos θ

Tool number one, for values. Theta, the Greek letter, is our angle. Often the question gives you tan theta, and asks for a fraction where each term, top and bottom, has exactly one sine theta or one cos theta. Okay? Now, we know only tan theta. So how can we turn the whole fraction into tan theta? Divide the top and the bottom by cos theta, yes, exactly! Sine theta over cos theta becomes tan theta, and cos theta over cos theta becomes one. The fraction keeps its value, because the top and the bottom are divided by the same thing. Right? And one check before we divide: cos theta must not be zero. Here it is safe: tan theta is given, so it is defined, and that needs cos theta not zero. The equals sign with a slash through it means not equal to. Clear?

1:52Tool 2: the bound on a cos theta + b sin theta

Tool number two, for bounds, from Chapter two. Take a times cos theta, plus b times sine theta. We can write it as the square root of a squared plus b squared, times cos of theta minus phi. The long tick is the square root sign, the small two up high means squared, and phi, another Greek letter, is a fixed angle with tan phi equal to b over a. Okay? Now, cos of theta minus phi is just a cosine. So how big can it ever be, and how small? From minus one to one, exactly! So the whole expression stays between minus root of a squared plus b squared, and plus root of a squared plus b squared. The sign with a bar under it means less than or equal to. Got it? Let us test it on our puzzle. In three sine theta plus four cos theta, the cos coefficient is four and the sine coefficient is three. Four squared plus three squared is twenty five, and its square root is five. So it lies from minus five to five, and the square brackets mean both end points are included. Right? And one condition from the course: the bound is reached only if the angle we need is allowed. When theta can be any angle, there is no problem; on a restricted interval, first find where theta minus phi lies. With me?

3:11Tool 3: allied angles

Tool number three, from Chapter one: allied angles. Take an angle n pi over two, plus or minus theta, where n is an integer, pi over two is ninety degrees, and the sign with a plus over a minus means plus or minus. When n is odd, sine and cos swap; that is what the double headed arrow means. Okay? And the sign comes from the original function, in the quadrant where the whole angle lies, treating theta as acute. Clear? Now try it on sine of theta plus pi over two, the angle in our puzzle. Here n is one, which is odd. So what does it become, and with which sign? It becomes plus cos theta, that is correct! Sine switches to cos, and theta plus pi over two lies in the second quadrant, where sine, the original function, is positive. Got it? And the course graphs on the right show the same thing. The solid blue curve is sine x, the dashed orange curve is cos x, and the cos curve is just the sine curve shifted left by pi over two. The green graph is tan x, which we do not need today.

4:18Example 1: tan theta = 2, find (2 sin + 3 cos)/(4 sin - cos)

Right? Now, Example one, an easy problem from Set one. If tan theta equals two, find two sine theta plus three cos theta, over four sine theta minus cos theta. Pause the video and try it first. Have you got an answer? Okay, let's check. Many students draw a right triangle here, to find sine theta and cos theta. But tan theta is positive in the first quadrant and also in the third, where sine and cos are both negative. So, do we need to know which quadrant theta is in? No, we do not, and that is the beauty of tool one! Every term has one sine or one cos, so we divide the top and the bottom by cos theta. It is not zero, because tan theta is defined. Okay? Now the fraction is two tan theta plus three, over four tan theta minus one. Put tan theta equal to two; the dot means times. Two times two, plus three, is seven, and four times two, minus one, is also seven. Seven over seven gives one. Clear? And if theta is in the third quadrant, sine and cos both change sign, so the top and the bottom both flip, and the answer is still one. So the answer is one, the course answer, in just two lines!

5:39Example 2: max of 3 sin + 4 cos + 5 sin(theta + pi/2)

Now, remember our puzzle? Here it is as Example two, a medium problem from Set one. Find the greatest value of three sine theta, plus four cos theta, plus five sine of theta plus pi over two. Pause the video and try it first. Have you got an answer? Okay, let's check. Many students say that the first two terms are at most five, the last term is at most five, so the answer is ten. Is that right? Not so fast, we will settle ten at the end. First, combine the terms. By tool three, five sine of theta plus pi over two is five cos theta. Right? Now add the cos terms: four cos theta plus five cos theta is nine cos theta. So the whole expression is just three sine theta plus nine cos theta. Okay? Now it has the shape of tool two, with a equal to nine and b equal to three. So what is the greatest value? The square root of a squared plus b squared, yes, exactly! Nine squared is eighty one, and three squared is nine, so we get the square root of ninety, which is three root ten, the course answer. Theta can be any angle here, so this bound is really reached. And the wavy equals sign means approximately: three root ten is about nine point four nine. Clear? So ten is never reached, and here is why. The last term is five only when cos theta is one, and at that angle the first two terms give just four. So the sum is four plus five, which is nine, not ten. So our puzzle is solved: the two parts never peak at the same theta, so always combine first, then take the bound. Got it?

7:24Quick revision: divide by cos, root(a2+b2), combine first

So, quick revision. First, when tan theta is given, divide the top and the bottom by cos theta. Second, a times cos theta, plus b times sine theta, lies between minus and plus root of a squared plus b squared. Third, combine the terms first, then take the bound; never add separate maximum values. Now try the practice problems on this page, and then the rest of Set one, at one to two minutes each. All the best!