Hello TeriyakiDonnQ Originally Posted by

**TeriyakiDonnQ** Hiii~

I don't understand the double angle identity~ Can anyone explain to me? :3

So i'm given

cos81sin81

and i don't know what to do with it. I have the answers and it says it's 1/2sin162.

I don't understand how you get the answer D;

Annd how do you do

2cos^2 3.4-1

the answer to that one is cos6.8

I don't get it at all. ); can anyone explain?

thaanks :]

The 'double-angle' identities you're referring to tell you how to find the sine and cosine of an angle if you know the sine and cosine of the angle half its size. Here's the one for sine:$\displaystyle \sin 2A = 2\sin A \cos A$ ...(1)

So this means, for example, that:$\displaystyle \sin 70^o = 2\sin35^o\cos35^o$

$\displaystyle \sin 8^o =2\sin 4^o\cos4^o$

... and so on.

Now if you divide both sides of (1) by 2, you'll get:$\displaystyle \sin A \cos A =\tfrac12\sin2A$

Can you see, then, that if $\displaystyle A = 81^o$, this will give:$\displaystyle \sin81^o\cos81^o=\tfrac12\sin162^o$

With the formula for $\displaystyle \cos2A$, there's more than one possibility. You can start with:$\displaystyle \cos2A = \cos^2A-\sin^2A$

but then, if you use the identity $\displaystyle \cos^2A +\sin^2A = 1$ (which I'm sure you know), you can write, alternatively:$\displaystyle \cos2A = 2\cos^2A-1$

or:$\displaystyle \cos2A = 1-2\sin^2A$

Which one of these three you use depends on the circumstances, but they are all correct.

You are given the expression $\displaystyle 2\cos^23.4 - 1$. Well, that's just the RHS of the second of these three cosine formulae with $\displaystyle A = 3.4$. (If you simply didn't spot this, that's because you haven't had enough practice yet!)

So, using the second formula with $\displaystyle A = 3.4$, we get:$\displaystyle 2\cos^23.4 - 1 = \cos 2\times3.4=\cos6.8$

Grandad