Finding the derivative of sinus and cosinus. Trigonometric identities

Question

How can we see that $$\sin(x+h)-\sin(x)=2\sin\left(\frac h2\right)\cos\left(x+\frac h2\right)$$

How can we see that $$\cos(x+h)-\cos(x)=-2\sin\left(\frac h2\right)\sin\left(x+\frac h2\right)$$

Do these identities have a name?

Answer 1

You asked for a name: Looks like an application of the reverse of the Prosthaphaeresis identities, or sum-to-product identities.

$$ \sin a − \sin b = 2 \cos\frac{a + b}{2} \sin\frac{a − b}{2} \\ \cos a − \cos b = −2 \sin\frac{a + b}{2} \sin\frac{a − b}{2} $$ using $a = x + h$ and $b = x$.

Another link here.

I usually use the complex definitions of $\sin$ and $\cos$ to proof, but you probably want an elementary one.

Note there are four identities in total.

$$ \sin a + \sin b = 2 \sin\frac{a + b}{2} \cos\frac{a − b}{2} \\ \cos a + \cos b = 2 \cos\frac{a + b}{2} \cos\frac{a − b}{2} $$

which would yield $$ \sin(x+h) + \sin(x) = 2 \sin\left(x+\frac{h}{2}\right) \cos\left(\frac{h}{2}\right) \\ \cos(x+h) + \cos(x) = 2 \cos\left(x+\frac{h}{2}\right) \cos\left(\frac{h}{2}\right) $$

Answer 2

For any $a,b$, we have the well known "addition theorems" $$ \sin(a+b) = \sin a \cos b + \sin b \cos a$$ and $$ \sin(a-b) = \sin a \cos b - \sin b \cos a $$ Subtracting these two equations, we get $$ \sin(a+b) - \sin(a-b) = 2\sin b \cos a $$

For the cosine, we have $$ \cos(a+b) = \cos a \cos b - \sin a \sin b $$ and $$ \cos(a-b) = \cos a \cos b + \sin a \sin b $$ Subtraction again $$ \cos(a+b) - \cos(a-b) = -2\sin a\sin b$$ Now let $b = \frac h2$, $a = x+\frac h2$.