How do I prove that 2 sin ((C+D)/2) cos ((C-D)/2) = sin C+sin D?

Nov 20, 2017

$\text{see explanation}$

Explanation:

$\text{using the "color(blue)"trigonometric identities}$

•color(white)(x)sin(A+B)=sinAcosB+cosAsinB

•color(white)(x)sin(A-B)=sinAcosB-cosAsinB

$\text{Adding the 2 equations gives}$

$\sin \left(A + B\right) + \sin \left(A - B\right) = 2 \sin A \cos B$

$\text{Subtracting the 2 equations gives}$

$\sin \left(A + B\right) - \sin \left(A - B\right) = 2 \cos A \sin B$

$\text{let "C=A+B" and } D = A - B$

$\Rightarrow A = \frac{C + D}{2} \text{ and } B = \frac{C - D}{2}$

$\Rightarrow \sin C + \sin D = 2 \sin \left(\frac{C + D}{2}\right) \cos \left(\frac{C - D}{2}\right)$

Nov 20, 2017

See the proof below

Explanation:

We need

$\sin \left(a + b\right) = \sin a \cos b + \sin b \cos a$

$\cos \left(a - b\right) = \cos a \cos b + \sin a \sin b$

${\sin}^{2} a + {\cos}^{2} a = 1$

$\sin 2 a = 2 \sin a \cos a$

Therefore,

$L H S = 2 \sin \left(\frac{C + D}{2}\right) \cos \left(\frac{C - D}{2}\right)$

$= 2 \left(\sin \left(\frac{C}{2}\right) \cos \left(\frac{D}{2}\right) + \sin \left(\frac{D}{2}\right) \cos \left(\frac{C}{2}\right)\right) \left(\cos \left(\frac{C}{2}\right) \cos \left(\frac{D}{2}\right) + \sin \left(\frac{C}{2}\right) \sin \left(\frac{D}{2}\right)\right)$

$= 2 \left(\sin \left(\frac{C}{2}\right) \cos \left(\frac{C}{2}\right) {\cos}^{2} \left(\frac{D}{2}\right) + \sin \left(\frac{D}{2}\right) \cos \left(\frac{D}{2}\right) {\sin}^{2} \left(\frac{C}{2}\right) + \sin \left(\frac{D}{2}\right) \cos \left(\frac{D}{2}\right) {\cos}^{2} \left(\frac{C}{2}\right) + \sin \left(\frac{C}{2}\right) \cos \left(\frac{C}{2}\right) {\sin}^{2} \left(\frac{D}{2}\right)\right)$

$= 2 \cdot \frac{1}{2} \left(\sin C {\cos}^{2} \left(\frac{D}{2}\right) + \sin D {\sin}^{2} \left(\frac{C}{2}\right) + \sin D {\cos}^{2} \left(\frac{C}{2}\right) + \sin C {\sin}^{2} \left(\frac{D}{2}\right)\right)$

$= \sin C \left({\cos}^{2} \left(\frac{D}{2}\right) + {\sin}^{2} \left(\frac{D}{2}\right)\right) + \sin D \left({\sin}^{2} \left(\frac{C}{2}\right) + {\cos}^{2} \left(\frac{C}{2}\right)\right)$

$= \sin C + \sin D$

$= R H S$

$Q E D$