MHB Dc 8t14 product to sum indentity

[karush]

4Use the power to sum formula to simplify the expression

$\frac{\sin\left({3\theta}\right)+\sin\left({5\theta}\right)}
{\cos\left({3\theta}\right)+\cos\left({5\theta}\right)}$

The answer is $\tan(4\theta)$

$$\sin\left({3\theta}\right)+\sin\left({5\theta}\right)
=2\sin\left({\frac{3\theta+5\theta}{2}}\right)\cos\left({\frac{3\theta-5\theta}{2 }}\right)$$

$$\cos\left({3\theta}\right)+\cos\left({5\theta}\right)
=2\cos\left({\frac{3\theta+5\theta}{2}}\right)\cos\left({\frac{3\theta-5\theta}{2 }}\right)$$

Hopefully I'm going in the right direction... But couldn't get the answer earlier

 

[MarkFL]

We are given:

$$\frac{\sin(3\theta)+\sin(5\theta)}{\cos(3\theta)+\cos(5\theta)}$$

We have the following two applicable sum to product formulas:

$$\sin(x)+\sin(y)=2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$$

$$\cos(x)+\cos(y)=2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$$

So, how can we use these to rewrite the given expression?

edit: I see you've now edited your post and have the correct identities. So, using what you've written, we would have:

$$\frac{\sin(3\theta)+\sin(5\theta)}{\cos(3\theta)+\cos(5\theta)}=\frac{2\sin\left(\frac{3\theta+5\theta}{2}\right)\cos\left(\frac{3\theta-5\theta}{2}\right)}{2\cos\left(\frac{3\theta+5\theta}{2}\right)\cos\left(\frac{3\theta-5\theta}{2}\right)}$$

Now, just simplify. :)

 

[karush]

$$\frac{2\sin\left({4\theta}\right)\cos\left({-\theta}\right)}
{2\cos\left({4\theta}\right)\cos\left({-\theta}\right)}
=\frac{\sin\left({4\theta}\right)}{\cos\left({4\theta}\right)}=\tan\left({4\theta}\right)$$

 

[MarkFL]

$$\frac{2\sin\left({4\theta}\right)\cos\left({-\theta}\right)}
{2\cos\left({4\theta}\right)\cos\left({-\theta}\right)}
=\frac{\sin\left({4\theta}\right)}{\cos\left({4\theta}\right)}=\tan\left({4\theta}\right)$$

Well, $3\theta-5\theta=-2\theta$ and the using the identity $\cos(-x)=\cos(x)$ you would have $\cos(2\theta)$ in the numerator and denominator, but I wouldn't even bother with that...I would write:

$$\frac{\sin(3\theta)+\sin(5\theta)}{\cos(3\theta)+\cos(5\theta)}=\frac{\cancel{2}\sin\left(\frac{3\theta+5\theta}{2}\right)\cancel{\cos\left(\frac{3\theta-5\theta}{2}\right)}}{\cancel{2}\cos\left(\frac{3\theta+5\theta}{2}\right)\cancel{\cos\left(\frac{3\theta-5\theta}{2}\right)}}$$

$$\frac{\sin(3\theta)+\sin(5\theta)}{\cos(3\theta)+\cos(5\theta)}=\frac{\sin\left(\frac{8\theta}{2}\right)}{\cos\left(\frac{8\theta}{2}\right)}$$

$$\frac{\sin(3\theta)+\sin(5\theta)}{\cos(3\theta)+\cos(5\theta)}=\frac{\sin\left(4\theta\right)}{\cos\left(4\theta\right)}=\tan(4\theta)$$

 

[karush]

Well, that was helpful, didn't see that, Nice thing about MHB, learn shortcuts

How do you do the strike through? (Cool)​

 

[MarkFL]

Well, that was helpful, didn't see that, Nice thing about MHB, learn shortcuts

How do you do the strike through? (Cool)​

You can use the \cancel command...there is also the \xcancel{} command which will x out a factor. Both of these can be found in our Qucik $\LaTeX$ element in the "Algebra" section.