When you solve equations, you start with an equation and you manipulate it until you get to a solution. You don't know what that solution will be, when you start, but you know the techniques that will get you there.
When you prove identities, on the other hand, you start with an equation and... you know you're supposed to do something (what?) to one of the sides (which?) to get to the other side (how?).
Content Continues Below
If you remember your high-school geometry, you might recall being flummoxed by the "fill in the blanks" proofs you were supposed to do. They gave you the "givens" at the top, whatever it was that you were supposed to prove at the bottom, and a few blank lines in between. What the heck was supposed to go in those lines?
Affiliate
Advertisement
A good technique, I've found, is to cover up the blank lines, leaving just the givens and the thing to prove. Then think, "Okay, what do these givens make me think of? What do they feel like?" And put something down on the paper (that was covering over the blank lines). Then see where that would take you. Soon enough, you'd have proven what you were supposed to prove. And, almost always, when you pulled that paper aside, you'd note that you just so happened to take as many steps as they'd expected you to, and now you could fill in the blanks.
Why did that work? In large part, it was by design. The authors knew what they'd taught you already, and they'd given you a proof that was just a very small step forward from that. There may have been more than one way to prove it, but they were consciously steering you, so the steps you came up with were the steps that they'd intended you to take. And, once students see this, they are able to get over that psychological block.
The point I'd like to make here is that those proofs were completed by a process of, well, trying stuff. If something doesn't seem to be working, then backing up and trying something else. Otherwise, keep going. You can do this; you do know how. You just need to be willing to try stuff, and have faith in yourself.
I can't tell which side is more complicated, but I do see a difference of squares on the lhs, so I think I'll start there.
sin4(x) − cos4(x)
= [sin2(x) + cos2(x)][sin2(x) − cos2(x)]
The first factor, sin2(x) + cos2(x), is always equal to 1, so I can ignore it. This leaves me with:
sin2(x) − cos2(x)
Hmm... I'm not seeing much of anything here. But I do know, glancing back at the rhs of the identity, that I need more sines and fewer cosines. I think I'll try using the Pythagorean identity that simplified that first factor, but in a slightly different form. If sin2(x) + cos2(x) = 1, then cos2(x) = 1 − sin2(x), and:
sin2(x) − cos2(x)
= sin2(x) − [1 − sin2(x)]
= sin2(x) − 1 + sin2(x)
= 2sin2(x) − 1
Ah! That's what I needed; I started on the lhs, and I've now arrived at the rhs. For my hand-in work, I'll put it all together:
sin4(x) − cos4(x)
= [sin2(x) + cos2(x)][sin2(x) − cos2(x)]
= [1][sin2(x) − cos2(x)]
= sin2(x) − cos2(x)
= sin2(x) − [1 − sin2(x)]
= sin2(x) − 1 + sin2(x)
= sin2(x) + sin2(x) − 1
= 2sin2(x) − 1
(By the way, there is no mathematical significance to the square brackets; I just wanted to use grouping symbols that wouldn't get lost in all the parentheses.)
Content Continues Below
(1 − cos2(α))(1 + cos2(α)) = 2sin2(α) − sin4(α)
I think I'll start by multiplying out the lhs:
1 − cos2(α) + cos2(α) − cos4(α)
= 1 − cos4(α)
That doesn't seem to have gotten me anywhere. What if I apply the Pythagorean identity to that first factor? Then I'll get:
(1 − cos2(α))(1 + cos2(α))
= sin2(α)[1 + cos2(α)]
Hmm... That doesn't seem to have helped, either. Okay, what happens if I work on the other side? I can factor a squared sine out of the two terms:
2sin2(α) − sin4(α)
= sin2(α)[2 − sin2(α)]
If I break the 2 into two 1's, I can use that same Pythagorean identity again to turn one of the 1's into sines and cosines. (I think I'm detecting a theme here.)
sin2(α)[1 − sin2(α) + 1]
= sin2(α)[1 − sin2(α) + sin2(α) + cos2(α)]
= sin2(α)[1 + cos2(α)]
Wait a minute! That's the same thing I ended up with on the lhs! Aha!
While what I've done so far is not a proof, I have managed to get the two sides to meet in the middle. And sometimes that seems to be the only way to do the proof for some identities: work on the two sides until they meet in the middle, and then write something that looks like magic. I'm going to start with the lhs, work down to where the two sides meet, and then work up the rhs until I get back to the original identity:
(1 − cos2(α))(1 + cos2(α))
= sin2(α)[1 + cos2(α)]
= sin2(α)[1 + cos2(α) − sin2(α) + sin2(α)]
= sin2(α)[1 − sin2(α) + sin2(α) + cos2(α)]
= sin2(α)[1 − sin2(α) + 1]
= sin2(α)[2 − sin2(α)]
= 2sin2(α) − sin4(α)
Affiliate
This, so you know, is how the textbook authors come up with those magical proofs, where you wonder how on earth they ever came up with one or another of the steps: they worked both sides until they made them match, and then rewrote their steps, working down one side to the matching step, and then back up with other side.
There will be times you'll need to do this. But, for your hand-in work, make sure you do the work the right way: Work down one side to the meeting place, and then hop over to the other side and work back up to the starting place. Only by showing your steps in that way will your proof be valid. But, hey — now you can do magic, too! 😜
sin2(θ)sec2(θ) + sin2(θ)csc2(θ) = sec2(θ)
Clearly, the lhs is the more complicated side, so I'll start there, and will convert everything to sines and cosines:
The first fraction simplifies to the tangent, and the second fraction simplifies to 1.
What I'm left with is one of the Pythagorean identities:
tan2(θ) + 1 = sec2(θ)
...and that's what I needed to end up with. Putting it all together:
In addition to the "working on the more complicated side" and the "converting to sines and cosines" tricks, there is one other trick that hardly ever comes up, but it would probably be wise to familiarize yourself with it....
URL: https://www.purplemath.com/modules/proving2.htm
© 2024 Purplemath, Inc. All right reserved. Web Design by