Question

Which of the following are identities?

Answer 1

\[A) sinx + \sin5x = \tan3x\]
\[B) (sinx + cosx)^2 = 1 + \sin2x\]
\[C) \cos^23x - \sin^23x = \cos6x\]
\[D) \sin^6x + \cos^6x = 4 - 3\sin^22x\]

Answer 2

I know B is NOT an identity. The others are eluding me :(

Answer 3

plug in random numbers and check

Answer 4

But for identities, aren't you supposed to solve them some other way? My lesson said NOT to plug in... because even if certain values check out, it doesn't make it an identity necessarily...

Answer 5

example


\(\sin^6(x) + \cos^6(x) = 4 - 3\sin^2(2x)\)

plug in \(\pi\)

\(\sin^6(\pi) + \cos^6(\pi) \overset{?}{=} 4 - 3\sin^2(2\pi)\\0-1\overset{?}{=}4-3*0\)

Answer 6

we are just trying to eliminate some for now

Answer 7

if you plug in a bunch of numbers and they are they same on both sides, then most likely they are identities and then we move on and prove them

Answer 8

to show something is NOT an identity, it is enough to show that they are not equal for some value in the domain.

Answer 9

Oh okay. Is sin^6(x) the same thing as 6sin(x) ?

Answer 10

no

Answer 11

these are all different things

\(6\sin(x)\\
\sin^6(x)\\
\sin(6x)\)

Answer 12

Oh... how did you get sin^6(pi) = 0 then? :P

Answer 13

\(\sin^6(\pi)=(\sin(\pi))^6=0^6=0\)

Answer 14

Gotcha. So D is not an identity right?

Answer 15

correct

Answer 16

sinx+sin5x=tan3x
sin(0) + sin5(0) = tan3(0)
0 + 0 = 0
Umm so it worked? I think

Answer 17

sinx+sin5x=tan3x
sin(pi/3) + sin5(pi/3) = tan3(pi/3)
sqrt3/3 + -sqrt3/3 = 0
Yes?

Answer 18

So A is likely an identity.. now how do we prove it?

Answer 19

You're confident B is NOT an identity? :3
Interesting...

Answer 20

Actually, I think I messed up on that... I was thinking (sinx + cosx)^2 = 1 but that's not true, is it?

Answer 21

Naw, because,\[\large\rm (\sin x+\cos x)^2\ne \sin^2x +\cos^2x\]You get a little bit more than that to work with.

Answer 22

(sinx+cosx)2=1+sin2x
sin^2x + 2cosxsinx + cos^2x = 1 + 2sinxcosx
1 + 2cosxsinx = 2sinxcosx
B is an identity. :D :D :D

Answer 23

What about A?

Answer 24

Prove that A is likely an identity?
Oh boy... hmm

Answer 25

I don't think there is any way that A could be an identity.
Try plugging in "problem values".
Values that might cause a problem for tangent.

Answer 26

I tried 0 and pi/3, both worked. So I'm thinking it is an identity... but I'm not sure. I'll try some more values.

Answer 27

tan(pi/2) is the bad place, ya?

So maybe ... pi/(3*2) is what we need, ya?
3(pi/6) = pi/2

Answer 28

I say "ya" too much... sorry, I'll work on that -_- no I won't.

Answer 29

I tried pi/4 and got
sqrt2/2 - sqrt/2 = -1
Which is not true... so A is out

Answer 30

I say "haha" too much xD don't worry about it

Answer 31

http://www.wolframalpha.com/input/?i=sinx+%2B+%5Csin5x+%3D+%5Ctan3x

Not an identity, use some values from the graph to show it

Answer 32

What specifically am I looking for on the graph?

Answer 33

https://www.desmos.com/calculator/m0x78tnrqu
Notice that the functions overlap at 0 and pi/3.
LOL, man that was some bad luck on your first two plug ins :)

Answer 34

http://www.wolframalpha.com/input/?i=sinx+%2B+sin(5x)+%3D+tan(3x)

Answer 35

Look for any places where the functions aren't intersecting, which... is like 99.9% of them

Answer 36

Best thing imo is to pick completely random values, not common ones.

Test crap like x= 0.1376 or some random junk.

Answer 37

They are never gonna give you an equation which has the exact solution of 0.1376. EVER.

Answer 38

Ya, but then she's gotta bust out one of those calculator doohickeys...
and who wants to do that ;( really..

Answer 39

Well yeah but if you use crap like pi/3, you have to use your brain, and who wants to do that.

Answer 40

You have to remember things... see the video for why that is bad
https://youtu.be/0FThgw8mXWo

Answer 41

haha xD you guys

Answer 42

I'll always remember the unit circle values... MEMORY DELETED.

Answer 43

Lolol.
A: no
B: yes
C: ?
D: ?

Answer 44

C is just your Cosine Double Angle, ya?\[\large\rm \cos^2(stuff)-\sin^2(stuff)=\cos(double~stuff~oreos)\]

Answer 45

XD

Answer 46

What.
lol

Answer 47

\[\large\rm \cos^2(stuff)-\sin^2(stuff)=\cos(2~stuff)\]That's your Cosine Double Angle formula, ya?\[\large\rm \cos^2(3x)-\sin^2(3x)=\cos(2\cdot3x)\]

Answer 48

C is true, see zeps oreo thing above. http://www.wolframalpha.com/input/?i=%5Ccos%5E2(3x)+-+%5Csin%5E2(3x)+%3D+%5Ccos(6x)


D is colossally false http://www.wolframalpha.com/input/?i=%5Csin%5E6x+%2B+%5Ccos%5E6x+%3D+4+-+3%5Csin%5E2(2x)

Answer 49

I tested x= 0.1376 and it checked out :D for C

Answer 50

Ohmygosh thank you both so much! This has been helpful

Answer 51

yay team

Answer 52

I learned a lot about... oreos... and stuff

Answer 53

OREOS DO NOT WORK THAT WAY! GOODNIGHT!
https://youtu.be/PmDVHs-juPo

Answer 54

Hahaha xD agent is in a funny mood today

Answer 55

Mr. pokerface has a sense of humor :)

Answer 56

I amused myself when i realized it's pretty much a fact that you'd never be given an equation with the solution x=0.1376.

But @zepdrix's double stuff oreos bit got me, and then Futurama clips did the rest.

Answer 57

"you'd never be given an equation with the solution x=0.1376"
^hilarious
lolol

Answer 58

Haha you could make an algebraic equation with that, but i'd think it pretty difficult to make a trig equation with that solution (without inverse trig functions, cos that'd be cheating)

But algebraic equations for x=0.1376 would be, as Morbo says of his family, belligerent and numerous https://youtu.be/aQPakERjJrI

Answer 59

You just love that don't you xD

Answer 60

Now I respect @zepdrix. I think he's a good man, but quite frankly, I agree with everything he just said! https://youtu.be/Ll3iyvbsRDM

Answer 61

Futurama huh? 0_o
interesting

Answer 62

Ha, that was a good one.

Answer 63

@zepdrix if you don't watch Futurama, I will lose all respect for you and punch you. https://youtu.be/cVejweBfLc0

Answer 64

Crap now I gotta go fix my YouTube Search History -_-
I don't want a bunch of Futurama crap showing up on my recommended lol

Answer 65

And why not?!

Answer 66

Oooh

Answer 67

I sense a fight brewing.