So the webpage title would be: How do you show that sin i*theta = i*sinh(theta)?

In summary, sinh(theta) = .5*[{cos(theta) + isin(theta)}+e^i - {cos(-theta) + isin(-theta)} + e^i] and sin(i*theta) = 1/i*sinh(i*theta)
  • #1
StephenD420
100
0
I need to show that sin i*theta= i* sinh(theta).
where sinh(theta) = .5[e^theta - e^(-theta)]
and cos(theta) = .5[e^theta + e^(-theta)]
and e^(i*theta) = cos(theta) + isin(theta)
if I start with the formula sinh(theta) = .5[e^theta - e^(-theta)]
and plug in e^(i*theta) = cos(theta) + isin(theta)
I get

sinh(theta) = .5*[{cos(theta) + isin(theta)}+e^i - {cos(-theta) + isin(-theta)} + e^i]

since cos(-theta) = cos(theta) and sin(-theta) = -sin(theta)

sinh(theta) = .5*2*i*sin(theta)
or
sinh(theta) = i*sin(theta)

now how do I go from here to
sin(i*theta) = i*sinh(theta)

I know I am almost there I just need a little last nudge.
Thanks
Stephen
 
Last edited:
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  • #2
Are you sure it isnt..

[tex]sinh(i \theta) = isin\theta[/tex]??
 
  • #3
nope..

sin(i*theta) = i*sinh(theta)

Attached is the question as given by the professor
 

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  • #4
if I start with the formula sinh(theta) = .5[e^theta - e^(-theta)]

Okay, fair enough.

and plug in e^(i*theta) = cos(theta) + isin(theta)
I get

sinh(theta) = .5*[{cos(theta) + isin(theta)}+e^i - {cos(-theta) + isin(-theta)} + e^i]

This isn't true. Try finding [itex] \sin(i\theta)[/itex] first with Euler's formula, and then compare the result to what you were given for [itex]\sinh(\theta)[/itex].
 
  • #5
How would I find \sin(i\theta)

from Eulers formula
e^itheta = cos(theta) +isin(theta)

so if theta = itheta then would
e^itheta = e^-theta = cos(itheta) +i*sin(itheta)
so
sin(i*theta) = (e^-theta -cos(itheta))/i

how does this help?
 
  • #6
Alright, I'm going to show you a little bit of wizardry. We know that

[tex]e^{i\theta}=\cos(\theta) + i \sin(\theta) \; \; \; [1][/tex]

Now let [itex]\theta = -\theta[/itex] so that:

[tex]e^{i(-\theta)}=\cos(-\theta) + i \sin(-\theta)[/tex]

Using the facts that cosine is even and sine is odd (which you already knew):

[tex]e^{i(-\theta)}=\cos(\theta) - i \sin(\theta) \; \; \; [2][/tex]

What happens if you subtract [2] from [1]?
 
  • #7
StephenD420 said:
I need to show that sin i*theta= i* sinh(theta).
where sinh(theta) = .5[e^theta - e^(-theta)]
and cosh(theta) = .5[e^theta + e^(-theta)]
and e^(i*theta) = cos(theta) + isin(theta)

if I start with the formula sinh(theta) = .5[e^theta - e^(-theta)]
and plug in e^(i*theta) = cos(theta) + isin(theta)
I get

sinh(theta) = .5*[{cos(theta) + isin(theta)}+e^i - {cos(-theta) + isin(-theta)} + e^i]
...

I know I am almost there I just need a little last nudge.
Thanks
Stephen
I don't see how you get
sinh(θ) = .5*[{cos(θ) + i sin(θ)}+ei - {cos(-θ) + i sin(-θ)} + ei]
...​

That's equivalent to [itex]\sinh(\theta)=.5\{e^{i\theta}+e^{i}-e^{-i\theta}+e^{i} \}[/itex] which is definitely not true.

BTW: It is true that sin(iθ) = i sinh(θ) .
 
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  • #8
ok..Screwdriver

I did that and got
.5(e^(itheta)-e^(-itheta) = isin(theta)
so
sinh(theta) = i sin(theta)

now how do I get from here to
sin(i*theta) = i*sinh(theta)
 
  • #9
You've determined that [itex]\sin(\theta)=\frac{1}{2i}(e^{i\theta}-e^{i\theta})[/itex]. Here is where you should sub in [itex]\theta = i\theta[/itex].
 
  • #10
ok if you plug in i*theta into sin(theta) = 1/i *sinh(\theta) you get sin(i*theta) = 1/i * sinh(i*theta)

How do you go from this to sin(i*theta) = i*sinh(theta)
 
  • #11
Screwdriver said:
You've determined that [itex]\sin(\theta)=\frac{1}{2i}(e^{i\theta}-e^{i\theta})[/itex]. Here is where you should sub in [itex]\theta = i\theta[/itex].
That should be [itex]\sin(\theta)=\frac{1}{2i}(e^{i\theta}-e^{-i\theta})\,.[/itex] What you had was equivalent to zero.
 
  • #12
SammyS said:
That should be [itex]\sin(\theta)=\frac{1}{2i}(e^{i\theta}-e^{-i\theta})\,.[/itex] What you had was equivalent to zero.

Oops, thanks for catching that typo :redface:

StephenD420 said:
ok if you plug in i*theta into sin(theta) = 1/i *sinh(\theta) you get sin(i*theta) = 1/i * sinh(i*theta)

How do you go from this to sin(i*theta) = i*sinh(theta)

You should be subbing in [itex]i\theta[/itex] into [itex]\sin(\theta)=\frac{1}{2i}(e^{i\theta} - e^{i(-\theta)})[/itex]. In other words, simplify the following:

[tex]\sin(i\theta)=\frac{1}{2i}(e^{i(i\theta)} - e^{i(-i\theta)})[/tex]

And then compare it to the definition of [itex]\sinh(\theta)[/itex] in terms of exponential functions (given in the question.)
 

1. What are basic trig identities?

Basic trig identities are equations that relate the values of trigonometric functions to each other. These identities are used to simplify and solve trigonometric expressions and equations.

2. What is the most important trig identity?

The most important trig identity is the Pythagorean identity, which states that sin^2x + cos^2x = 1. This identity is used extensively in trigonometry and is the basis for many other identities.

3. How do you prove a trig identity?

To prove a trig identity, you need to manipulate one side of the equation using algebraic and trigonometric identities until it is equal to the other side. This process is called verifying the identity.

4. What are the most commonly used trig identities?

Some of the most commonly used trig identities include the reciprocal identities (cscx = 1/sinx, secx = 1/cosx, cotx = 1/tanx), the quotient identities (tanx = sinx/cosx, cotx = cosx/sinx), and the double angle identities (sin2x = 2sinxcosx, cos2x = cos^2x - sin^2x).

5. How are trig identities used in real life?

Trig identities are used in many real-life applications, such as in engineering, physics, and navigation. They are used to calculate distances, angles, and heights in various situations, such as in surveying land, building structures, and designing machines.

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