# Basic Trig Identity question

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)

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:

Are you sure it isnt..

$$sinh(i \theta) = isin\theta$$??

nope..

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

Attached is the question as given by the professor

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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 $\sin(i\theta)$ first with Euler's formula, and then compare the result to what you were given for $\sinh(\theta)$.

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???

Alright, I'm going to show you a little bit of wizardry. We know that

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

Now let $\theta = -\theta$ so that:

$$e^{i(-\theta)}=\cos(-\theta) + i \sin(-\theta)$$

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

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

What happens if you subtract [2] from [1]?

SammyS
Staff Emeritus
Homework Helper
Gold Member
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)

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 $\sinh(\theta)=.5\{e^{i\theta}+e^{i}-e^{-i\theta}+e^{i} \}$ which is definitely not true.

BTW: It is true that sin(iθ) = i sinh(θ) .

Last edited:
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)

You've determined that $\sin(\theta)=\frac{1}{2i}(e^{i\theta}-e^{i\theta})$. Here is where you should sub in $\theta = i\theta$.

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)

SammyS
Staff Emeritus
Homework Helper
Gold Member
You've determined that $\sin(\theta)=\frac{1}{2i}(e^{i\theta}-e^{i\theta})$. Here is where you should sub in $\theta = i\theta$.
That should be $\sin(\theta)=\frac{1}{2i}(e^{i\theta}-e^{-i\theta})\,.$ What you had was equivalent to zero.

That should be $\sin(\theta)=\frac{1}{2i}(e^{i\theta}-e^{-i\theta})\,.$ What you had was equivalent to zero.

Oops, thanks for catching that typo

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 $i\theta$ into $\sin(\theta)=\frac{1}{2i}(e^{i\theta} - e^{i(-\theta)})$. In other words, simplify the following:

$$\sin(i\theta)=\frac{1}{2i}(e^{i(i\theta)} - e^{i(-i\theta)})$$

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