Question

I give medals!
i need to prove this identity:

Answer 1

\[\cos (\frac{ \pi }{ 4 } +x)+\cos (\frac{ \pi }{ 4 }-x)= \sqrt{2} cosx \]

Answer 2

@jhannybean

Answer 3

Hm, there's an identity that falls with cosines and the addition or subtraction of angles. do you know what it is?

Answer 4

yes one second

Answer 5

cos(A+B) = (cosA)(cosB) + (sinA)(sinB) with this formula, it could be + or - based on the way it needs to be used

Answer 6

Yes :)

Answer 7

So we could use the cosine subtraction rule first \[\cos(A-B) = \cos(A)\cos(B)+\sin(A)\sin(B)\]\[\rm cos\left(\pi/4 -x\right) = \cos(\pi/4)\cos(x) +\sin(\pi/4)\sin(x)\]This completes the first part of the identity.\[\boxed{\cos(\pi/4)\cos(x) + \sin(\pi/4)\sin(x)} +\cos\left(\pi/4) + x\right) = \sqrt{2}\cos(x)\]

Answer 8

okay so far so good

Answer 9

Now evaluate \(\cos\left(\frac{\pi}{4}\right)\) and \(\sin\left(\frac{\pi}{4}\right)\)

Answer 10

and that is the \[\frac{ \sqrt{2} }{ 2 }\]

Answer 11

Or otherwise written as \(\frac{1}{\sqrt{2}}\)

Answer 12

so i now need to make that 1 disappear right?

Answer 13

So from this step\[\rightarrow \boxed{\cos(\pi/4)\cos(x) + \sin(\pi/4)\sin(x)} +\cos\left(\pi/4) + x\right) = \sqrt{2}\cos(x)\] we can substitute the values we've just found and simplify. \[\frac{\cos(x)}{\sqrt{2}}+\frac{\sin(x)}{\sqrt{2}} +\cos(\pi/4 +x) =\sqrt{2}\cos(x)\]

Answer 14

and like you mentioned earlier, we have our identity for the addition of angles within the cosine function. \[\cos(A+B) =\cos(A)\cos(B) -\sin(A)\sin(B)\]\[\cos(\pi/4 + x) = \cos(\pi/4)\cos(x)-\sin(\pi/4)\sin(x)\]

Answer 15

okay, im just reading through what youve put up. its starting to piece together and make more sense

Answer 16

So you end up with the same result. \[\frac{\cos(x)}{\sqrt{2}}+\frac{\sin(x)}{\sqrt{2}}+\boxed{\cos(\pi/4)\cos(x)-\sin(\pi/4)\sin(x)} =\sqrt{2}\cos(x)\]

Answer 17

my original problem only involved cos. Since sin is in the answer, is that a valid answer?

Answer 18

also, basically we started with only cos, but in order to get to the answer, we had to involve sin

Answer 19

is that right?

Answer 20

Yes, the sine function only came into the picture because of the cosine addition and subtraction rules.

Answer 21

ohh okay, thank you :)

Answer 22

\[\rightarrow \frac{\cos(x)}{\sqrt{2}}+\frac{\sin(x)}{\sqrt{2}}+\boxed{\cos(\pi/4)\cos(x)-\sin(\pi/4)\sin(x)} =\sqrt{2}\cos(x)\]\[\frac{\cos(x)}{\sqrt{2}} +\frac{\sin(x)}{\sqrt{2}}+\frac{\cos(x)}{\sqrt{2}}-\frac{\sin(x)}{\sqrt{2}}=\sqrt{2}\cos(x)\]

Answer 23

You see how the sine functions cancel eachother out?

Answer 24

yes, because of their signs (- and +), right?

Answer 25

Yes :)

Answer 26

Now we're left with \[\frac{2\cos(x)}{\sqrt{2}}=\sqrt{2}\cos(x)\]and we know \(2 =\sqrt{2}\cdot \sqrt{2}\) so the fraction disappears, and we're left with\[\sqrt{2}\cos(x) = \sqrt{2}\cos(x)\]:D!

Answer 27

i love you. thank you. so much.
could you double check just one more, i promise? i already solved it, can i check with you (under a different post) to confirm that i did it correctly? youre extremely smart. ill give you another medal (a third medal is pretty much the equivalent of a trophy ;) )

Answer 28

wow that really clarified everything

Answer 29

haha thank you! Sure, i you liek, rework the problem and I shall point out your mistakes. Keep this tab open so you can cross reference my work to yours :P

Answer 30

okay, you sure? i just want to credit you where due

Answer 31

I'm sure :)

Answer 32

this is difficult to put on here, im going to upload the math i did for it lol

Answer 33

Alllrighty :) Take a picture of your work on paper! :)

Answer 34

i am :) just waiting for it to download to my slow computer -_-

Answer 35

okay almost don. sorry it wsant sending to my email. 2 seconds

Answer 36

sorry its illegible, can you make out the problem?

Answer 37

Think you did something wrong there, it should come out to \[\boxed{\frac{\sqrt{3}}{2}}\]

Answer 38

Yeah, I felt like I did. Do you mind explaining how you got that?

Answer 39

Working on it, burning myself out a bit, lol.

Answer 40

Okay, it looks like you were making it a little harder than it actually was.

Answer 41

Okay, no problem, sorry. Btw if it looks like I sign out and sign back in or something its because my computers being a jerk at the moment.

Answer 42

That should be my middle name when it comes to math actually. I always make it harder than it is lol

Answer 43

\[\cos^2(15) - \sin^2(15)\] This identity states \[\cos^2(x)-\sin^2(x) \implies 2\cos^2(x) -1 \implies 1-2\sin^2(x) \implies \cos(2x)\]

Answer 44

So there are various forms you can use in solving for whatever you're solving for.

Answer 45

I personally like using \(2\sin^2(x)-1\) when I'm given a problem that relates to the double angle formula involving cosine, so I used that.

Answer 46

You've been insanely helpful this entire time, so I'm with you on that version. it's a-ok by me.

Answer 47

\[\cos^2(15) -\sin^2(15)\] Let's change the degree to radians. You do this by :\[15~ \text{deg} \cdot \frac{\pi}{180 ~\text{deg}} = \frac{\pi}{12} \]

Answer 48

wow this guy is smart

Answer 49

Sorry to interrupt... But I found an even shorter method
Using double argument
cos^2 (x) - sin^2 (x) =cos(2x)

And from your problem, you see that x=15
So all you have to find is
cos(2x15)
=cos(30)
And using the unit circle or whatever,
You know cos(30)= sqrt3/2

Answer 50

Yeah that works too :)

Answer 51

Hmm, okay, thanks yeah as long as I understand the method behind the madness

Answer 52

I was gonna say jhannybean, you've stuck by me on this stuff, and I understand the way you were explaining it, so you can complete it your way if you want as well

Answer 53

So we have : \[\cos^2\left(\frac{\pi}{12}\right) -\sin^2\left(\frac{\pi}{12}\right) \]\[=2\cos^2\left(\pi/12\right) -1\]\[=\frac{\sqrt{3}}{2}\]

Answer 54

I'm sorry I cannot help you with any more problems after this, I'm totally burnt out and a bit exhausted.

Answer 55

I totally understand, I appreciate all you've done, thanks

Answer 56

What I suggest you do is take a gigantic index card and write out all the formulas you are learning in class. When you do your homework, instead of looking through your book for the identities, you will have your flashcards with you.

Answer 57

That way not only will you lern the formulas, you can memorize and apply them later on with ease. Atleast that's how I went about doing my Trigonometry homework. :)

Answer 58

learn*

Answer 59

That makes sense, I'll def do it! Thanks a million. A thousand gold trophies for you!!!! THOUSANDS!

Answer 60

Haha thanks. Good luck with learning the rest of your identities! :-)

Answer 61

Thank you!! Take care!