Educational

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,

If sin 2A = λ sin 2B, then write the value of (λ + 1)/(λ − 1) If \( \sin2A=\lambda\sin2B \), then write the value of \( \dfrac{\lambda+1}{\lambda-1} \) Solution: Given, \[ \sin2A=\lambda\sin2B \] \[ \lambda=\frac{\sin2A}{\sin2B} \] Therefore, \[ \frac{\lambda+1}{\lambda-1} = \frac{ \frac{\sin2A}{\sin2B}+1 }{ \frac{\sin2A}{\sin2B}-1 } \] \[ = \frac{\sin2A+\sin2B}{\sin2A-\sin2B} \] Using identities, \[ \sin

Write the value of sin(π/15) sin(4π/15) sin(3π/10) Write the value of \( \sin\frac{\pi}{15}\sin\frac{4\pi}{15}\sin\frac{3\pi}{10} \) Solution: Using identity, \[ \sin x\sin(60^\circ-x)\sin(60^\circ+x) = \frac14\sin3x \] Here, \[ x=\frac{\pi}{15}=12^\circ \] Then, \[ 60^\circ-x=48^\circ=\frac{4\pi}{15} \] and \[ 60^\circ+x=72^\circ=\frac{2\pi}{5} \] Therefore, \[ \sin\frac{\pi}{15}\sin\frac{4\pi}{15}\sin\frac{2\pi}{5} = \frac14\sin\frac{\pi}{5} \] Since, \[ \sin\frac{3\pi}{10} = \sin\left(\frac{\pi}{2}-\frac{\pi}{5}\right) = \cos\frac{\pi}{5} \] Using exact value relation, \[ \sin18^\circ\sin42^\circ\sin54^\circ=\frac18

If A + B = π/3 and cos A + cos B = 1, then find cos((A − B)/2) If \( A+B=\frac{\pi}{3} \) and \( \cos A+\cos B=1 \), then find the value of \( \cos\frac{A-B}{2} \) Solution: Using identity, \[ \cos A+\cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} \] \[ 1 = 2\cos\frac{\pi/3}{2}\cos\frac{A-B}{2} \] \[ = 2\cos\frac{\pi}{6}\cos\frac{A-B}{2}