Question

Please help!
Prove the following identity
(sin x + cos x)2 = 1 + 2 sin x cos x

Answer 1

Know how proofs like this work in general?

Answer 2

sin^2x + 2sinxcosx + cos^x = 1 + 2sinxcosx
(sin^2 x + cos^2 x = 1) so..
1 + 2sinxcosx = 1 + 2sinxcosx

Answer 3

it was cos^2 x in the first part, sorry

Answer 4

no certain way @e.mccormick

Answer 5

can you rewrite it? @maferlicious thanks!

Answer 6

ohhhh never mind i got it @maferlicious

Answer 7

you mean instead of cos^x its cos^2x?

Answer 8

In general, you make both sides look the same by working on only one side. That is what maferlicious did, but some might argue with the number of steps (no shortcuts in proofs.)

Answer 9

okay okay can you tell me which side I should try to change? Thats what im having problems with

Answer 10

It is up to how you choose to start. It has an = sign, so it can be flipped.

\((\sin x + \cos x)^2 = 1 + 2 \sin x \cos x\)
is the same thing as:
\((1 + 2 \sin x \cos x=\sin x + \cos x)^2\)

So you can start on the \((\sin x + \cos x)^2\) side and make it into \(1 + 2 \sin x \cos x\) or the other way around, but you are not allowed to work on both sides at once.

This means your only math tools are multiply by 1, add 0, or replace with something that has the same value.

Answer 11

Oops... missed moving the (. LOL. But you get the point of how I just flipped sides...

Answer 12

yea I do thank you!

Answer 13

Now, adding 0 can be more complex than just adding 0. +1-1 could be added and that is 0. Or, because \(\sin^2\theta+\cos^2\theta=1\) you can say that \(\sin^2\theta+\cos^2\theta-1=0\) so you can add \(\sin^2\theta+\cos^2\theta-1\) to one side of an equation and it is the same as adding 0!

Multiplying by 1 is also more complex that people think. \(\dfrac{\cos\theta}{\cos\theta}\) ois the same thing as 1. Anyting over itself is.

As for replacing things, well, \(\tan\theta=\dfrac{\sin\theta}{\cos\theta}\) is one example. You can replace one side with the other. Same with anything else on your trig identities sheet.

Answer 14

Other tip with trig proofs that I always used and helped me: although you can't technically work off of both sides, it often helped me to work off both sides and then have them meet in the middle and just rewrite the steps from one side in the opposite order on the other side to make it proper
It can often help, because a lot of times where you get stuck is when something isn't obviously simplifying and you're actually making it more complicated in order to get to the other side, and it will be more apparent what you need to do by looking at the other side too :)

Answer 15

Yes, what scorch said is right. And if you do it on scratch paper, nobody will know! Just write the answer out properly.

Answer 16

Makes sense! thanks! I wanted to now if when simplifing can you have more than 1 function? For example: sec(theta)+tan(theta)

Answer 17

@e.mccormick ?

Answer 18

Well, simplified can be pretty complex at times. The best thing there is to do what you can to start (sometimes there are easy replacments or cancels,) then put everything in sine and cosine and simplify it as much as possible that way. The end should be pretty basic.

Answer 19

for example: 1/cos(θ) +tan(θ) =(1 + sin(θ))/cos(θ) For this I keep getting a simplified version of sec(θ)+tan(θ)

Answer 20

@e.mccormick

Answer 21

then when I work it out by putting everything on one side then making it equal to 0 I get the simplified version of 0

Answer 22

On, 1/cos(θ) +tan(θ) =(1 + sin(θ))/cos(θ) are you trying to simplify or prove that is true?

Answer 23

yes

Answer 24

you have to prove the identity

Answer 25

OK. This is one of those times where older math comes into play.

Basic fraction rules. To add two fractions, they must have the same what?

Answer 26

denominator so change them all to cosθ?

Answer 27

by changing tan too sin/cos

Answer 28

Yes. And tangent will play nice if... hehe, you got it before I finished hinting!

Answer 29

haha! So then the finished product would just be 1+sin/cos?

Answer 30

Yep.

1/cos(θ) +tan(θ) =(1 + sin(θ))/cos(θ) \(\implies\)
1/cos(θ) +sin(θ)/cos(θ) =(1 + sin(θ))/cos(θ) \(\implies\)
(1 +sin(θ))/cos(θ) =(1 + sin(θ))/cos(θ)

Trig is where you begin to really see what all the foundational math is for. All those fraction rules, how multiplying by 1 means nothing changes, etc. In calculus you see even more of it! So if you have any weak point in math at all, it is best to practice that area because trig and calc can find it and make past weaknesses into present difficulties.

Answer 31

perfect! Thats the answer I got! Thanks so much for your help!

Answer 32

np. Have fun!