Question

Help trig proofs

Answer 1

The equation \[e^{i \pi}+1=0 \] is sometimes called Eulers equations. One of his identies is \[e^{i \pi} = cosx +isinx\]

Answer 2

Verify Eulers equation using Eulers identity stated above. HInt: let x=pi

Answer 3

so this mean \[e^{i \pi}=\cos \pi+isin \pi\]

Answer 4

cospi = 1 sin pi =0 \[e^{e \pi} + 1=0\]

Answer 5

@ganeshie8

Answer 6

\[\large \color{brown}{e}^{\color{red}{i}\color{purple}{\pi}} +\color{blue}{1}=\color{magenta}{0}\]

do you see why this equation is so beautiful ?

Answer 7

Yes, but this is my first time seeing this. When exactly is this used?

Answer 8

It is from complex analysis. Notice that it has \(5\) most used math symbols : \[e,i,\pi,1,0\]

Answer 9

It does indeed. Could you assist with the next parts. It builds upon this

Answer 10

\[\large e^{i\color{red}{x}} = \cos \color{red}{x} + i\sin \color{red}{x}\]

you're using that right ?

Answer 11

Yes :)

Answer 12

\[\large e^{i\color{red}{\pi}} = \cos \color{red}{\pi} + i\sin \color{red}{\pi} = -1 + i0 = -1\]

Answer 13

Next part of the question says verify the next two Euler identities.
\[sinx=\frac{ e^{ix} - e^{-ix} }{ 2i}\] and \[cosx=\frac{ e^{ix}+e^{-ix} }{ 2 }\]

Answer 14

As you said, everythign follows from the definition :
\[\large e^{i\color{red}{x}} = \cos \color{red}{x} + i\sin \color{red}{x}\]

Answer 15

so we can replace e^ix with cosx+isinx?

Answer 16

cept in the case of -i ?

Answer 17

\(\large e^{i\color{red}{x}} = \cos \color{red}{x} + i\sin \color{red}{x}\)

\(\large e^{i(\color{red}{-x})} = \cos (\color{red}{-x}) + i\sin (\color{red}{-x})\)

Answer 18

Add them, what do you get ?

Answer 19

\[\sin \pi= \frac{ (\cos \pi+isin \pi)-(\cos -\pi+isin -\pi) }{ 2i}\]

Answer 20

forget about \(\pi\), we're done with that..

Answer 21

But it wants us to verify the identity

Answer 22

also it gave a hint to use what we got in part a

Answer 23

Oh ok.. you want to verify the identity at \(x=\pi\) is it ?

Answer 24

then it looks good :) keep going ...

Answer 25

Oh wait "subtitute identity from part a in the right side of the idtentity and simplify. So i beleive you are right

Answer 26

\[\sin x= \frac{ (\cos x+isin x)-(\cos -x+isin -x) }{ 2i}\]

like that ?

Answer 27

Yes sorry about miss reading

Answer 28

So we can multiply 2i to the other side?

Answer 29

so we get 2isinx= (2isinx)?

Answer 30

since the cosines get subtracted and the -x for sin is pulled out in front making it isinx+isinx?

Answer 31

Yes

Answer 32

numerator becoems
isinx+isinx = 2isinx

Answer 33

Does this mean we verified it?

Answer 34

have we ?

Answer 35

No because we arent proving one side is equal to the other?

Answer 36

we are solving it

Answer 37

Right hand side :

\(\large \frac{ (\cos x+i\sin x)-(\cos(-x)+i\sin (-x)) }{ 2i}\)

\(\large = \frac{ 2i\sin x }{ 2i}\)

\(\large =\sin x\)

Answer 38

which is same as the left hand side

so we have verified that identity...

Answer 39

I thought verifying and proving were different things oops

Answer 40

In general they are different things,
in our case they seem to be same yeah :)

Answer 41

Its fun how math works like this :p

Answer 42

So the next part is Use Eulers identities to write sin(1+i) in the standard form a+bi. Round to 3 decimal places

Answer 43

\[\sin x=\frac{ e^{ix} - e^{-ix} }{ 2i}\]

plugin \(x = 1+i\)

Answer 44

So is this using the one we just verified then?

Answer 45

Yes that is what i thought I see the pattern

Answer 46

Also, notice that many exponent identities work just fine with complex numbers too :
\[e^{a+ib} = e^a*e^{ib}\]