MHB Sum or difference formula (sin, cos, and tan)

[Taryn1]

So I'm supposed to find the exact values of the sine, cosine, and tangent of an angle by using a sum or difference formula ( i.e. sin(x+y)=sin(x)cos(y)+cos(x)sin(y) ), but this is the angle I was given: ${-13\pi}/{12}$. How do I use a sum or difference formula to get the sin, cos, and tan of that?

 

[MarkFL]

I would first add $2\pi$ to get:

$$-\frac{13}{12}\pi+2\pi=\frac{11}{12}\pi$$

And then write:

$$\frac{11}{12}\pi=\frac{1}{2}\pi+\frac{1}{4}\pi+\frac{1}{6}\pi$$

Now you can use the angle-sum formulas. :)

 

[Prove It]

I would first add $2\pi$ to get:

$$-\frac{13}{12}\pi+2\pi=\frac{11}{12}\pi$$

And then write:

$$\frac{11}{12}\pi=\frac{1}{2}\pi+\frac{1}{4}\pi+\frac{1}{6}\pi$$

Now you can use the angle-sum formulas. :)

Or even just $\displaystyle \begin{align*} \frac{3\pi}{4} + \frac{\pi}{6} \end{align*}$ or $\displaystyle \begin{align*} \frac{\pi}{4} + \frac{2\pi}{3} \end{align*}$ to avoid multiple uses of the compound angle formulae...

 

[Taryn1]

Thanks for your help! That makes more sense now.