This is a known trigonometry identity.  Whenever you are adding two sine functions, you can plug $\displaystyle x$ and $\displaystyle y$ into the formula to solve for this sum

We begin by considering our formula for the sum of two sines

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

We will let$\displaystyle x = \frac{\pi}{3} + x$ and $\displaystyle y = \frac{\pi}{3} - x$ and plug these values into our formula.

$\displaystyle sin(\frac{\pi}{3} +x) + sin(\frac{\pi}{3} - x) = 2sin\frac{(\frac{\pi}{3} + x) + (\frac{\pi}{3} - x)}{2}cos\frac{(\frac{\pi}{3} + x) - (\frac{\pi}{3} - x)}{2}$

$\displaystyle sin(\frac{\pi}{3} + x) + sin(\frac{\pi}{3} - x) = 2sin\frac{\frac{2\pi}{3}}{2}cos\frac{2x}{2}$

$\displaystyle sin(\frac{\pi}{3} + x) + sin(\frac{\pi}{3} - x) = 2sin\frac{\pi}{3}cosx$

$\displaystyle sin(\frac{\pi}{3} + x) + sin(\frac{\pi}{3} - x) = 2(\frac{\sqrt{3}}{2})(1)$

$\displaystyle sin(\frac{\pi}{3} + x) + sin(\frac{\pi}{3} - x) = \sqrt{3}$

We begin by considering the formula for the differences of two cosines.

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

We will let  $\displaystyle x = 2+x$  and  $\displaystyle y = 2 - x$.  Proceed by plugging these values into the formula.

$\displaystyle cos(2+x) - cos(2 -x) = -2sin\frac{(2+x) + (2 - x)}{2}sin\frac{(2+x) - (2-x)}{2}$

$\displaystyle cos(2+x) - cos(2 -x) = -2sin\frac{4}{2}sin\frac{2x}{2}$

$\displaystyle cos(2+x) - cos(2 -x) = -2sin(2)sin(x)$

This is a known trigonometry identity and has been proven to be true.  It is often helpful to solve for the quantity within a cosine function when there are unknowns or if the quantity needs to be simplified

To solve this problem we must use the identity

$\displaystyle sin(\alpha + \beta) = sin(\alpha)cos(\beta) + cos(\alpha)sin(\beta)$.  We will let $\displaystyle \alpha = \frac{\pi}{4}$ and  $\displaystyle \beta = \frac{\pi}{6}$.

$\displaystyle sin(\frac{\pi}{4} + \frac{\pi}{6}) = sin(\frac{\pi}{4})cos(\frac{\pi}{6}) + cos(\frac{\pi}{4})sin(\frac{\pi}{6})$

$\displaystyle sin(\frac{\pi}{4} + \frac{\pi}{6}) = (\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) + (\frac{\sqrt{2}}{2})(\frac{1}{2})$

$\displaystyle sin(\frac{\pi}{4} + \frac{\pi}{6}) =\frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{2}$

$\displaystyle sin(\frac{\pi}{4} + \frac{\pi}{6}) =\frac{\sqrt{6} + 2\sqrt{2}}{4}$

This answer is false.  $\displaystyle cos(\alpha - \beta)$ is not the same as $\displaystyle cos(\alpha) - cos(\beta)$.

For example, say $\displaystyle \alpha = \frac{2\pi}{3}$ and $\displaystyle \beta = \frac {\pi}{3}$

$\displaystyle cos(\frac{2\pi}{3} - \frac{\pi}{3}) = cos(\frac{\pi}{3}) = \frac{1}{2}$

$\displaystyle cos(\frac{2\pi}{3}) - cos(\frac{\pi}{3} = -\frac{1}{2} - \frac{1}{2} = -1$

And so the correct identity to use for this is

$\displaystyle cos(\alpha - \beta) = cos(\alpha)cos(\beta) +sin(\alpha)sin(\beta)$

The correct identity to use for this kind of problem is

$\displaystyle cos(\alpha + \beta) = cos(\alpha)cos(\beta) - sin(\alpha)sin(\beta)$. We will let $\displaystyle \alpha = 45 = \frac{\pi}{4}$  and   $\displaystyle \beta = 30 = \frac{\pi}{6}$.

$\displaystyle cos(45 + 30) = cos(45)cos(30) - sin(45)sin(30)$

$\displaystyle cos(45 + 30) = cos(\frac{\pi}{4})cos(\frac{\pi}{6}) - sin(\frac{\pi}{4})sin(\frac{\pi}{6})$

$\displaystyle cos(45 + 30) = (\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) - (\frac{\sqrt{2}}{2})(\frac{1}{2})$

$\displaystyle cos(45 + 30) = \frac{\sqrt{6}}{2} - \frac{\sqrt{2}}{2}$

$\displaystyle cos(45 + 30) = \frac{\sqrt{6} - 2\sqrt{2}}{4}$

This is a known trigonometry identity and has been proven to be true.  It is often helpful to solve for the quantity within a cosine function when there are unknowns or if the quantity needs to be simplified

This formula is able to be derived directly from the identities for the sum and difference of cosines, .

   $\displaystyle cos(\alpha + \beta) = cos(\alpha)cos(\beta) - sin(\alpha)sin(\beta)$

$\displaystyle +cos(\alpha - \beta) = cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta)$

$\displaystyle cos(\alpha + \beta) + cos(\alpha - \beta) = 2cos(\alpha)cos(\beta)$

$\displaystyle \implies cos(\alpha)cos(\beta)=\frac{1}{2}[cos(\alpha + \beta) + cos(\alpha - \beta)]$

First, we must know the formula for the product of sines so that we know what we are searching for.  The formula for this identity is $\displaystyle sin(\alpha)sin(\beta) = \frac{1}{2}[cos(\alpha -\beta) - cos(\alpha + \beta)]$.  Using the known identities of the sum/difference of cosines, we are able to derive the product of sines in this way.  Sometimes it is helpful to be able to expand the product of trigonometric functions as sums.  It can either simplify a problem or allow you to visualize the function in a different way.