Ravi Kant Kumar

Prove that sin 50° + sin 10° = cos 20° Prove that: \[ \sin 50^\circ + \sin 10^\circ = \cos 20^\circ \] Solution Using the identity: \[ \sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2} \] Taking \[ A = 50^\circ,\qquad B = 10^\circ \] Then, \[ \sin 50^\circ + \sin 10^\circ

Prove that cos 100° + cos 20° = cos 40° Prove that: \[ \cos 100^\circ + \cos 20^\circ = \cos 40^\circ \] Solution Using the identity: \[ \cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2} \] Taking \[ A = 100^\circ,\qquad B = 20^\circ \] Then, \[ \cos 100^\circ + \cos 20^\circ

Prove that sin 38° + sin 22° = sin 82° Prove that: \[ \sin 38^\circ + \sin 22^\circ = \sin 82^\circ \] Solution Using the identity: \[ \sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2} \] Taking \[ A = 38^\circ,\qquad B = 22^\circ \] Then, \[ \sin 38^\circ + \sin 22^\circ

Express sin 2x + cos 4x as Product of Sines and Cosines Express the following as the product of sines and cosines: \[ \sin 2x + \cos 4x \] Solution The expression \[ \sin 2x + \cos 4x \] contains both sine and cosine functions, so first convert \(\cos 4x\) into sine form using: \[

Express cos 12x − cos 4x as Product of Sines and Cosines Express the following as the product of sines and cosines: \[ \cos 12x – \cos 4x \] Solution Using the identity: \[ \cos A – \cos B = -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2} \] Here, \[ A = 12x,\qquad B = 4x \]

Express cos 12x + cos 8x as Product of Sines and Cosines Express the following as the product of sines and cosines: \[ \cos 12x + \cos 8x \] Solution Using the identity: \[ \cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2} \] Here, \[ A = 12x,\qquad B = 8x \]

Express sin 5x − sin x as Product of Sines and Cosines Express the following as the product of sines and cosines: \[ \sin 5x – \sin x \] Solution Using the identity: \[ \sin A – \sin B = 2 \cos \frac{A+B}{2} \sin \frac{A-B}{2} \] Here, \[ A = 5x,\qquad B = x \]

Express sin 12x + sin 4x as Product of Sines and Cosines Express the following as the product of sines and cosines: \[ \sin 12x + \sin 4x \] Solution Using the identity: \[ \sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2} \] Here, \[ A = 12x,\qquad B = 4x \]

If cos(A + B) sin(C − D) = cos(A − B) sin(C + D), then find tan A tan B tan C If \( \cos(A+B)\sin(C-D)=\cos(A-B)\sin(C+D) \), then write the value of \( \tan A\tan B\tan C \) Solution: Using identities, \[ \cos(A+B)=\cos A\cos B-\sin A\sin B \] and \[ \cos(A-B)=\cos A\cos B+\sin A\sin B \]

Write the value of (sin A + sin 3A)/(cos A + cos 3A) Write the value of \( \dfrac{\sin A+\sin3A}{\cos A+\cos3A} \) Solution: Using identities, \[ \sin C+\sin D = 2\sin\frac{C+D}{2}\cos\frac{C-D}{2} \] \[ \sin A+\sin3A = 2\sin2A\cos A \] Also, \[ \cos C+\cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2} \] \[ \cos A+\cos3A = 2\cos2A\cos A \] Therefore,