MHB Secondary Identity Confirmation

[Dundee3]

Hey fellahs, got another whopper that's killing me.

$$1 - \cos2\theta + \cos8\theta - \cos10\theta=?$$

My objective here is to complete the identity, and my worksheet lists the correct solution as:

$$4\sin\theta\cos4\theta\sin5\theta$$

And once again I've had trouble beating this one. This is what I've conjured so far:

$$1 - (1 - 2\sin^2\theta) + (-2\sin((8\theta + 10\theta)/2)\sin((8\theta-10\theta)/2)$$

$$1 -1 + 2\sin^2\theta + 2\sin9\theta\sin\theta$$

$$2\sin^2\theta - 2\sin9\theta * -\sin\theta$$
And from this point I'm stumped =\

Any help would be awesome!
Thanks again, homies.

 

[MarkFL]

I would look at:

$$1-\cos(10\theta)=2\sin^2(5\theta)$$

$$\cos(8\theta)-\cos(2\theta)=-2\sin(5\theta)\sin(3\theta)$$

And now your original expression can be factored as:

$$2\sin(5\theta)\left(\sin(5\theta)-\sin(3\theta)\right)$$

Now, we know:

$$\sin(5\theta)-\sin(3\theta)=2\sin(\theta)\cos(4\theta)$$

Now to finish is fairly easy...:D

 

[Dundee3]

It makes perfect sense! Thank you so much!

I was never familiar with the identity:

$$1 - \cos10\theta = 2\sin^2(5\theta)$$

Thank you again!

 

[MarkFL]

It makes perfect sense! Thank you so much!

I was never familiar with the identity:

$$1 - \cos10\theta = 2\sin^2(5\theta)$$

Thank you again!

It is just a re-formulation of a double-angle identity for cosine:

$$\cos(2\theta)=1-2\sin^2(\theta)$$

 

[Greg]

$$1 -1 + 2\sin^2\theta + 2\sin9\theta\sin\theta$$

$$2\sin^2\theta+2\sin9\theta\sin\theta$$$$=2\sin\theta(\sin\theta+\sin9\theta)$$$$=2\sin\theta\cdot2\sin5\theta\cos4\theta$$$$=4\sin\theta\cos4\theta\sin5\theta$$