(i)

$\sin (A\pm B)=\sin A.\cos B\pm \cos A.\sin B$ 

(ii)

$\cos (A\pm B)=\cos A.\cos B\mp \sin A.\sin B$

(i)

$\sin C+\sin D=2\sin \frac{C+D}{2}.\cos \frac{C–D}{2}$ 

(ii)

$\sin C–\sin D=2\cos \frac{C+D}{2}.\sin \frac{C–D}{2}$ 

(iii)

$\cos C+\cos D=2\cos \frac{C+D}{2}.\cos \frac{C–D}{2}$ 

(iv)

$\cos C–\cos D=2\sin \frac{C+D}{2}.\sin \frac{D–C}{2}$

(i)

$\sin 2A=2\sin A.\cos A$ 

(ii)

$\cos 2A={\cos }^{2}A–{\sin }^{2}A=2{\cos }^{2}A–1=1–2{\sin }^{2}A$ 

(iii)

$\tan 2A=\frac{2\tan A}{1–{\tan }^{2}A}$

हल: हम जानते हैं कि 75° = 45° + 30°

$\sin (45°+30°)=\sin 45°\cos 30°+\cos 45°\sin 30°$ 

$=(\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2})+(\frac{1}{\sqrt{2}}\times \frac{1}{2})=\frac{\sqrt{3}+1}{2\sqrt{2}}$

हल: सूत्र:

$\tan 2A=\frac{2\tan A}{1–{\tan }^{2}A}$ 

$=\frac{2(1/3)}{1–(1/9)}=\frac{2/3}{8/9}=\frac{2}{3}\times \frac{9}{8}=\frac{3}{4}$

हल: C & D सूत्र का उपयोग करते हुए:

$\sin C+\sin D=2\sin \frac{C+D}{2}\cos \frac{C–D}{2}$ 

$=2\sin \frac{40+20}{2}\cos \frac{40–20}{2}=2\sin 30°\cos 10°$

$\cos (A+A)=\cos A\cos A–\sin A\sin A$cos 2A = cos² A – sin² A (इति सिद्धम)

हल: सूत्र:

$\cos C–\cos D=2\sin \frac{C+D}{2}\sin \frac{D–C}{2}$ 

$=2\sin \frac{100+20}{2}\sin \frac{20–100}{2}$ 

$=2\sin 60°\sin (–40°)=–2\sin 60°\sin 40°$