A CRUX MOVE is a trick that comes up over and over when doing math. If you worked lots of problems, you'd probably eventually notice them. However, I took the time to notice them for you and made a list and included some examples. Although my examples often come from calculus, the tricks are applicable in any area.
There's good news and bad news.
The good news is Crux Move #4 doesn't require much thinking. It's just a list of things to remember.
The bad news is it's trig identities. And memorizing trig identities is about exciting as sitting through your Aunt Edith's wedding when you're 14. Her Catholic wedding. In Savannah. In August. And all your friends are out at the lake blowing up stuff with the M-80s Donny Kerabatsos stole from his brother.
Trust me, fidgeting isn't going to make this any easier.
Yup. I said memorize. Not "know they exist" or "know where to find them." Memorize.
Why can't I just look them up?
That's a fair question. The answer comes in two parts. First, you may be in a situation where you can't look them up. An exam, for example. Or the GRE. Sometimes you may be given or allowed to bring a list of trig identities and other useful factoids. But not always (we weren't). And you don't want to get stuck memorizing assumed background material in the days leading up to a test.
More importantly, memorization brings internalization. There are uses of trig identities that don't announce themselves with a big flashing arrow (see examples below). Not obvious. Not conspicuous. Not self-evident. If a crux move is stuck in some dusty list you figured to consult as needed, you won't. Because you'll never know it was in there.
Internalization brings prowess. And, hey, bonus: knowing the ids off the top of your head brings speed. Off to the pub sooner.
That being said, I feel your pain. We got a huge list of these things back at the start of Calc-I (and were tested on them. As in: here's a blank sheet of paper write down as many as you can). Such collections tend to be overkill (the ones in my CRC go on for 20 pages). We can substantially prune this burden. Many of the identities are the same thing expressed in different ways. Pick the version you like best. And many identities are simply bozo weirdness you'll never use. That's a little trickier business, gazing into the future.
Still, I've kept an eye on the trig identities I have called upon with any regularity and made a list. There's seven categories (reduction, cofunction, Pythagorean, sum, difference, double angle, power-reducing) with an identity for sine, cosine, and tangent in most. I've also thrown in a few alternate forms at the end that are easily derivable. It goes without saying you apply these to secant, cosecant, and cotangent by working with the appropriate inverses.
Here they are. Print out. Stick up somewhere. I've included some gentle chiding in the title to encourage regular drill. Examples after.
E X A M P L E 1
Express [ 1 – cos(x) ] / sin(x) using a single trig function.
Applying, in turn, power-reducing and double-angle identities, we obtain
[ 1 – cos(x) ] / sin(x) = [ 2⋅sin2(x/2) ] / sin(x)
= [ 2 sin2(x/2) ] / [ 2⋅sin(x/2)⋅cos(x/2) ]
= sin(x/2) / cos(x/2)
= tan(x/2)
that is, [ 1 – cos(x) ] / sin(x) = tan(x/2). This identity is also worth remembering (which is why it's also on the list).
E X A M P L E 2
Find cos(179) ⋅ cos(1) – sin(179) ⋅ sin(1) w/o using a calculator (angles in degrees).
Using a product-to-sum rule:
cos(179) ⋅ cos(1) – sin(179)⋅sin(1) = cos(179+1)
= cos(180)
= –1
This is the standard ...you're stranded on an island and you need this trig expression to escape but a monkey ran off with your calculator problem that's in every textbook. The lesson? If you're ever stranded on an island, the first thing you should do is kill all the monkeys.
E X A M P L E 3
Integrate: ∫ cos(4x)⋅cos(3x) dx
Trig identities are the key to integrating trig functions (you did/will do a week or two of these in Calc-II, probably). This example demonstrates one flavor. Rewrite the integrand using a product-to-sum rule:
∫ cos(4x)⋅cos(3x) dx = ∫ 1/2 [ cos(7x) + cos(x) ] dx
= 1/2 ∫ cos(7x) dx + 1/2 ∫ cos(x) dx
and the rest of the solution is straightforward. This breed of trick works for any integrand of the form sin(Ax)⋅sin(Bx), cos(Ax)⋅cos(Bx) or sin(Ax)⋅cos(Bx).
E X A M P L E 4
Integrate: ∫ sin3(x)⋅cos2(x) dx
First, factor out a sin2(x):
∫ sin3(x)⋅cos2(x) dx = ∫ sin2(x)⋅cos2(x)⋅sin(x) dx
Now do a Pythagorean on sin2(x):
= ∫ [1 – cos2(x)]⋅cos2(x)⋅sin(x) dx
= ∫ cos2(x)⋅sin(x) dx – ∫ cos4(x)⋅sin(x) dx
= –1/3 cos3(x) + 1/5 cos5(x) + C
This example illustrates what I mean by internalization. If you simply paged through a list of trig identities looking for something that matched this integrand, you would get nowhere. Knowing the Pythagorean identity is what tips us off to bring out a factor of sin2(x) in the first place. If you know this trick, you might claim the factoring and subsequent solution is obvious. But it isn't, and I dare say it wasn't the first time you encountered such a beast.
E X A M P L E 5
Compute the Taylor series for sin(x) expanded about π/2.
Usually, we expand a Taylor series about zero because it simplifies things. Here, we aren't so lucky. You could drag out the Taylor series machinery and start calculating stuff. Or, applying the cofactor identity sin(x) = cos(π/2 – x) you could just substitute into the Taylor series for cos(x). Recall
cos(x) = 1 – x2 / 2! + x4 / 4! – x6 / 6! + ...
so:
sin(x) = 1 – [π/2 – x]2 / 2! + [π/2 – x]4 / 4! – ... [1]
For comparison, recall the Taylor series for sin(x) expanded about zero:
sin(x) = x – x3 / 3! + x5 / 5! – x7 / 7! + ... [2]
You should pause here and convince yourself that [1] and [2] give the same result for all x. What's the difference between the two? Nothing really, assuming you include enough terms. (If you truncate each after a couple of terms and plot the results using a graphing calculator, you'd notice [1] is more accurate around π/2 and [2] is more accurate around zero. Granted, that's not earth-shaking.) Rather, this example hints at a Larger Math Truth: make life easier by tweaking what you already know. In this case we got a series for sine by tweaking the series for cosine using a trig identity. We shall be seeing more of this LMT in future crux moves.
Previous move: Assume a Solution
Next move: Complete the Square
There's good news and bad news.
The good news is Crux Move #4 doesn't require much thinking. It's just a list of things to remember.
The bad news is it's trig identities. And memorizing trig identities is about exciting as sitting through your Aunt Edith's wedding when you're 14. Her Catholic wedding. In Savannah. In August. And all your friends are out at the lake blowing up stuff with the M-80s Donny Kerabatsos stole from his brother.
Trust me, fidgeting isn't going to make this any easier.
Crux Move #4
Apply a Trig Identity
Apply a Trig Identity
Yup. I said memorize. Not "know they exist" or "know where to find them." Memorize.
Why can't I just look them up?
That's a fair question. The answer comes in two parts. First, you may be in a situation where you can't look them up. An exam, for example. Or the GRE. Sometimes you may be given or allowed to bring a list of trig identities and other useful factoids. But not always (we weren't). And you don't want to get stuck memorizing assumed background material in the days leading up to a test.
More importantly, memorization brings internalization. There are uses of trig identities that don't announce themselves with a big flashing arrow (see examples below). Not obvious. Not conspicuous. Not self-evident. If a crux move is stuck in some dusty list you figured to consult as needed, you won't. Because you'll never know it was in there.
Internalization brings prowess. And, hey, bonus: knowing the ids off the top of your head brings speed. Off to the pub sooner.
That being said, I feel your pain. We got a huge list of these things back at the start of Calc-I (and were tested on them. As in: here's a blank sheet of paper write down as many as you can). Such collections tend to be overkill (the ones in my CRC go on for 20 pages). We can substantially prune this burden. Many of the identities are the same thing expressed in different ways. Pick the version you like best. And many identities are simply bozo weirdness you'll never use. That's a little trickier business, gazing into the future.
Still, I've kept an eye on the trig identities I have called upon with any regularity and made a list. There's seven categories (reduction, cofunction, Pythagorean, sum, difference, double angle, power-reducing) with an identity for sine, cosine, and tangent in most. I've also thrown in a few alternate forms at the end that are easily derivable. It goes without saying you apply these to secant, cosecant, and cotangent by working with the appropriate inverses.
Here they are. Print out. Stick up somewhere. I've included some gentle chiding in the title to encourage regular drill. Examples after.
LabKitty's List of Trig Identities
Memorize or suffer
___ reduction ___
sin(–x) = –sin(x)
cos(–x) = cos(x)
tan(–x) = –tan(x)
___ cofunction ___
sin(π/2 – x) = cos(x)
cos(π/2 – x) = sin(x)
tan(π/2 – x) = cot(x)
___ Pythagorean ___
sin2(x) + cos2(x) = 1
sec2(x) – tan2(x) = 1
___ sum ___
sin(x+y) = sin(x)⋅cos(y) + cos(x)⋅sin(y)
cos(x+y) = cos(x)⋅cos(y) – sin(x)⋅sin(y)
tan(x+y) = [ tan(x) + tan(y) ] / [ 1 – tan(x)⋅tan(y) ]
___ difference ___
sin(x–y) = sin(x)⋅cos(y) – cos(x)⋅sin(y)
cos(x–y) = cos(x)⋅cos(y) + sin(x)⋅sin(y)
tan(x–y) = [ tan(x) – tan(y) ] / [ 1 + tan(x)⋅tan(y) ]
___ double angle ___
sin(2x) = 2⋅sin(x)⋅cos(x)
cos(2x) = 2⋅cos2(x) – 1
tan(2x) = [ 2⋅tan(x) ] / [ 1 – tan2(x) ]
alternative cos forms
cos(2x) = 1 – 2⋅sin2(x)
cos(2x) = cos2(x) – sin2(x)
___ power-reducing ___
sin2(x) = [ 1 – cos(2x) ] / 2
cos2(x) = [ 1 + cos(2x) ] / 2<
tan2(x) = [ 1 – cos(2x) ] / [1 + cos(2x) ]
Bonus! -- Derivables and alternate forms
tan(x) = sin(2x) / [ 1 + cos(2x) ]
tan(x) = [ 1 – cos(2x) ] / sin(2x)
product-to-sum
cos(x)⋅cos(y) = 1/2 [ cos(x+y) + cos(x–y) ]
sin(x)⋅sin(y) = 1/2 [ cos(x–y) – cos(x_y) ]
sum-to-product
sin(x) + sin(y) = 2⋅sin([x+y]/2)⋅cos([x–y]/2)
sin(x) – sin(y) = 2⋅cos([x+y]/2)⋅cos([x–y]/2)
sin(x) + sin(y) = 2⋅cos([x+y]/2)⋅cos([x–y]/2)
sin(x) + sin(y) = –2⋅sin([x+y]/2)⋅sin([x–y]/2)
Memorize or suffer
___ reduction ___
sin(–x) = –sin(x)
cos(–x) = cos(x)
tan(–x) = –tan(x)
___ cofunction ___
sin(π/2 – x) = cos(x)
cos(π/2 – x) = sin(x)
tan(π/2 – x) = cot(x)
___ Pythagorean ___
sin2(x) + cos2(x) = 1
sec2(x) – tan2(x) = 1
___ sum ___
sin(x+y) = sin(x)⋅cos(y) + cos(x)⋅sin(y)
cos(x+y) = cos(x)⋅cos(y) – sin(x)⋅sin(y)
tan(x+y) = [ tan(x) + tan(y) ] / [ 1 – tan(x)⋅tan(y) ]
___ difference ___
sin(x–y) = sin(x)⋅cos(y) – cos(x)⋅sin(y)
cos(x–y) = cos(x)⋅cos(y) + sin(x)⋅sin(y)
tan(x–y) = [ tan(x) – tan(y) ] / [ 1 + tan(x)⋅tan(y) ]
___ double angle ___
sin(2x) = 2⋅sin(x)⋅cos(x)
cos(2x) = 2⋅cos2(x) – 1
tan(2x) = [ 2⋅tan(x) ] / [ 1 – tan2(x) ]
alternative cos forms
cos(2x) = 1 – 2⋅sin2(x)
cos(2x) = cos2(x) – sin2(x)
___ power-reducing ___
sin2(x) = [ 1 – cos(2x) ] / 2
cos2(x) = [ 1 + cos(2x) ] / 2<
tan2(x) = [ 1 – cos(2x) ] / [1 + cos(2x) ]
Bonus! -- Derivables and alternate forms
tan(x) = sin(2x) / [ 1 + cos(2x) ]
tan(x) = [ 1 – cos(2x) ] / sin(2x)
product-to-sum
cos(x)⋅cos(y) = 1/2 [ cos(x+y) + cos(x–y) ]
sin(x)⋅sin(y) = 1/2 [ cos(x–y) – cos(x_y) ]
sum-to-product
sin(x) + sin(y) = 2⋅sin([x+y]/2)⋅cos([x–y]/2)
sin(x) – sin(y) = 2⋅cos([x+y]/2)⋅cos([x–y]/2)
sin(x) + sin(y) = 2⋅cos([x+y]/2)⋅cos([x–y]/2)
sin(x) + sin(y) = –2⋅sin([x+y]/2)⋅sin([x–y]/2)
E X A M P L E 1
Express [ 1 – cos(x) ] / sin(x) using a single trig function.
Applying, in turn, power-reducing and double-angle identities, we obtain
[ 1 – cos(x) ] / sin(x) = [ 2⋅sin2(x/2) ] / sin(x)
= [ 2 sin2(x/2) ] / [ 2⋅sin(x/2)⋅cos(x/2) ]
= sin(x/2) / cos(x/2)
= tan(x/2)
that is, [ 1 – cos(x) ] / sin(x) = tan(x/2). This identity is also worth remembering (which is why it's also on the list).
E X A M P L E 2
Find cos(179) ⋅ cos(1) – sin(179) ⋅ sin(1) w/o using a calculator (angles in degrees).
Using a product-to-sum rule:
cos(179) ⋅ cos(1) – sin(179)⋅sin(1) = cos(179+1)
= cos(180)
= –1
This is the standard ...you're stranded on an island and you need this trig expression to escape but a monkey ran off with your calculator problem that's in every textbook. The lesson? If you're ever stranded on an island, the first thing you should do is kill all the monkeys.
E X A M P L E 3
Integrate: ∫ cos(4x)⋅cos(3x) dx
Trig identities are the key to integrating trig functions (you did/will do a week or two of these in Calc-II, probably). This example demonstrates one flavor. Rewrite the integrand using a product-to-sum rule:
∫ cos(4x)⋅cos(3x) dx = ∫ 1/2 [ cos(7x) + cos(x) ] dx
= 1/2 ∫ cos(7x) dx + 1/2 ∫ cos(x) dx
and the rest of the solution is straightforward. This breed of trick works for any integrand of the form sin(Ax)⋅sin(Bx), cos(Ax)⋅cos(Bx) or sin(Ax)⋅cos(Bx).
E X A M P L E 4
Integrate: ∫ sin3(x)⋅cos2(x) dx
First, factor out a sin2(x):
∫ sin3(x)⋅cos2(x) dx = ∫ sin2(x)⋅cos2(x)⋅sin(x) dx
Now do a Pythagorean on sin2(x):
= ∫ [1 – cos2(x)]⋅cos2(x)⋅sin(x) dx
= ∫ cos2(x)⋅sin(x) dx – ∫ cos4(x)⋅sin(x) dx
= –1/3 cos3(x) + 1/5 cos5(x) + C
This example illustrates what I mean by internalization. If you simply paged through a list of trig identities looking for something that matched this integrand, you would get nowhere. Knowing the Pythagorean identity is what tips us off to bring out a factor of sin2(x) in the first place. If you know this trick, you might claim the factoring and subsequent solution is obvious. But it isn't, and I dare say it wasn't the first time you encountered such a beast.
E X A M P L E 5
Compute the Taylor series for sin(x) expanded about π/2.
Usually, we expand a Taylor series about zero because it simplifies things. Here, we aren't so lucky. You could drag out the Taylor series machinery and start calculating stuff. Or, applying the cofactor identity sin(x) = cos(π/2 – x) you could just substitute into the Taylor series for cos(x). Recall
cos(x) = 1 – x2 / 2! + x4 / 4! – x6 / 6! + ...
so:
sin(x) = 1 – [π/2 – x]2 / 2! + [π/2 – x]4 / 4! – ... [1]
For comparison, recall the Taylor series for sin(x) expanded about zero:
sin(x) = x – x3 / 3! + x5 / 5! – x7 / 7! + ... [2]
You should pause here and convince yourself that [1] and [2] give the same result for all x. What's the difference between the two? Nothing really, assuming you include enough terms. (If you truncate each after a couple of terms and plot the results using a graphing calculator, you'd notice [1] is more accurate around π/2 and [2] is more accurate around zero. Granted, that's not earth-shaking.) Rather, this example hints at a Larger Math Truth: make life easier by tweaking what you already know. In this case we got a series for sine by tweaking the series for cosine using a trig identity. We shall be seeing more of this LMT in future crux moves.
Previous move: Assume a Solution
Next move: Complete the Square
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