Conditional Identities

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. In a triangle, tan A is one and tan B is two. Can you find tan C, with no angle given? It sounds impossible, right? But it takes just one line, because the three angles of a triangle always add up to pi. Today we learn conditional identities, and they come up often in JEE questions. So, let's begin.

0:27True only under a condition

So, a conditional identity is true only when the angles obey a condition. Our condition is that capital A plus B plus C equals pi. Pi is the Greek letter on screen, and pi radians is one hundred and eighty degrees, so this is exactly what the angles of a triangle do. Right? Move C to the other side, and capital A plus B equals pi minus C, as simple as that. This small move is the key to everything today, so remember it. Clear? Now take the sine and the cos; the round brackets hold the angle. Sine of A plus B equals sine C, no change there. But cos of A plus B equals minus cos C, with a minus sign. Remember that minus sign. With me? So let me ask you this: what is tan of A plus B? Yes, exactly, it is minus tan C! Tan is sine over cos, so the minus sign from the cos comes along. Right?

1:25Sum of the tangents equals their product

Now the most famous result. Open tan of A plus B with the compound angle formula: tan A plus tan B, over one minus tan A tan B; the long line is the fraction bar. That equals minus tan C, as we just saw. Clear? Multiply both sides by the bottom, and open the bracket. Tan A plus tan B equals minus tan C, plus tan A times tan B times tan C, right? Now bring tan C over to the left side. What do you get? Beautiful, the sum of the three tangents equals their product! Tan A plus tan B plus tan C equals tan A times tan B times tan C, okay? This is the result that solves our puzzle. But careful, every tan must be defined, so no angle can be an odd multiple of pi over two. In a triangle, that simply means no ninety degree angle, so in a right-angled triangle this identity cannot be used. Got it?

2:24Sines and cosines when A+B+C=π

Alright, now sines and cosines. Sine two A plus sine two B plus sine two C equals four times sine A times sine B times sine C, okay? We will prove this one in example two. The cos two A version has a number in front. Is it plus one or minus one? It is minus one, then minus four times cos A times cos B times cos C, okay? Both signs are minus, so don't write plus one by habit. Next, the single angles. Sine A plus sine B plus sine C equals four times cos of A over two, times cos of B over two, times cos of C over two. Sines on the left, cosines of the half angles on the right. Clear? And cos A plus cos B plus cos C equals one, plus four times sine of A over two, times sine of B over two, times sine of C over two. Now it swaps to sines of the half angles, with a plus one in front. With me?

3:23Pairs that add up to one

Now divide the tangent identity by tan A times tan B times tan C, all three together. Since cot is one over tan, we get cot A cot B, plus cot B cot C, plus cot C cot A, equals one. Clear? Now let me ask you this: what if the angles add up to pi over two, instead of pi? Then tan and cot swap roles! Tan of A plus B equals cot C, and the same steps give tan A tan B, plus tan B tan C, plus tan C tan A, equals one. Got it? And now cot A plus cot B plus cot C equals cot A times cot B times cot C, the same pattern as before. So pi gives the tangent sum, and pi over two gives the cot sum. Don't mix them up! Here is a neat connection. In a triangle, the half angles, capital A over two, B over two and C over two, add up to pi over two, right? So they obey the pi over two results. So tan of A over two times tan of B over two, plus the next two pairs made the same way, equals one. With me?

4:32Topper's trick: how to prove them

Now a trick that toppers use to prove these results. Step one: combine two terms with a sum to product formula. Okay? Step two: replace capital A plus B by pi minus C, so sine of A plus B becomes sine C, and cos of A plus B becomes minus cos C, right? For the tangent results, start from tan of A plus B equals minus tan C, as we did.

4:57Example 1: tan C in a triangle when tan A = 1, tan B = 2

Now, our puzzle from the start. In a triangle, tan A is one and tan B is two, and we want the value of tan C, with no angle given. Pause the video and try it first. Done? Okay, let's check. Let small t stand for tan C, the unknown. Since the sum of the tangents equals their product, one plus two plus t equals one times two times t; the small dot means times. Clear? So three plus t equals two t, which gives t equals three. Tan C is three, in one line, as promised! Right? Let us verify it, using tan of A plus B on the third line. One plus two, over one minus two, is three over minus one, which is minus three, and that is minus tan C, perfect. No angle here is ninety degrees, so the identity applies. Clear?

5:51Example 2: prove sin 2A+sin 2B+sin 2C = 4 sin A sin B sin C

Now a harder one, a proof. If capital A plus B plus C equals pi, prove that sine two A plus sine two B plus sine two C equals four times sine A times sine B times sine C, our first result. Pause the video and try it first, using the trick. Done? Okay, let's check. Combine the first two terms with sum to product: half of two A plus two B is A plus B, and half of two A minus two B is A minus B, right? So we get two sine of A plus B, times cos of A minus B, and sine of A plus B is just sine C, as we saw. Clear? Now the third term. Sine two C is two sine C cos C, and cos C is minus cos of A plus B, from our key step. Be very careful with that minus sign, because dropping it breaks the proof, okay? Now add the two lines. Two sine C comes out as a common factor, and the square brackets hold what is left: cos of A minus B, minus cos of A plus B, in that order. With me? So, what is cos of A minus B, minus cos of A plus B? Expand both with the compound angle formulas. Yes, exactly, it is two sine A sine B, because the cos A cos B terms cancel! So we get two sine C times two sine A sine B, which is four times sine A times sine B times sine C, exactly what we had to prove. Got it?

7:24Quick revision: tan sum = tan product, and the pi/2 case

So, quick revision. When capital A plus B plus C equals pi, sine of A plus B is sine C, and cos of A plus B is minus cos C, with the minus sign. The sum of the three tangents equals their product, when every tan is defined. And when the angles add up to pi over two, the tan products in pairs add up to one, and the cot sum equals the cot product. Now go and try the practice problems on this page. All the best!