Educational

Prove that (cos 3A + 2 cos 5A + cos 7A)/(cos A + 2 cos 3A + cos 5A) = cos 5A/cos 3A Prove that: \[ \frac{\cos 3A + 2\cos 5A + \cos 7A} {\cos A + 2\cos 3A + \cos 5A} = \frac{\cos 5A}{\cos 3A} \] Solution L.H.S. \[ = \frac{\cos 3A + 2\cos […]

Prove that (sin A + sin 3A + sin 5A)/(cos A + cos 3A + cos 5A) = tan 3A Prove that: \[ \frac{\sin A + \sin 3A + \sin 5A} {\cos A + \cos 3A + \cos 5A} = \tan 3A \] Solution L.H.S. \[ = \frac{\sin A + \sin 3A + \sin 5A}

Prove that (cos A + cos B)/(cos B – cos A) = cot(A + B)/2 cot(A – B)/2 Prove that: \[ \frac{\cos A + \cos B}{\cos B – \cos A} = \cot\frac{A+B}{2}\cot\frac{A-B}{2} \] Solution L.H.S. \[ = \frac{\cos A + \cos B}{\cos B – \cos A} \] Use identity: \[ \cos A + \cos B

Prove that (sin A + sin B)/(sin A − sin B) = tan((A+B)/2) cot((A−B)/2) Prove that: \[ \frac{\sin A+\sin B}{\sin A-\sin B} = \tan\frac{A+B}{2} \cot\frac{A-B}{2} \] Solution Consider the left-hand side: \[ \frac{\sin A+\sin B}{\sin A-\sin B} \] Using the identities: \[ \sin A+\sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} \] \[ \sin A-\sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2} \]

Prove that (sin A − sin B)/(cos A + cos B) = tan((A − B)/2) Prove that: \[ \frac{\sin A-\sin B}{\cos A+\cos B} = \tan\frac{A-B}{2} \] Solution Consider the left-hand side: \[ \frac{\sin A-\sin B}{\cos A+\cos B} \] Using the identities: \[ \sin A-\sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2} \] \[ \cos A+\cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} \]

Prove that (sin 9A − sin 7A)/(cos 7A − cos 9A) = cot 8A Prove that: \[ \frac{\sin 9A-\sin 7A}{\cos 7A-\cos 9A} = \cot 8A \] Solution Consider the left-hand side: \[ \frac{\sin 9A-\sin 7A}{\cos 7A-\cos 9A} \] Using the identity: \[ \sin A-\sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2} \] \[ \sin 9A-\sin 7A = 2\cos\frac{9A+7A}{2}\sin\frac{9A-7A}{2} \]

Prove that (sin A + sin 3A)/(cos A − cos 3A) = cot A Prove that: \[ \frac{\sin A+\sin 3A}{\cos A-\cos 3A} = \cot A \] Solution Consider the left-hand side: \[ \frac{\sin A+\sin 3A}{\cos A-\cos 3A} \] Using the identity: \[ \sin A+\sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} \] \[ \sin A+\sin 3A = 2\sin\frac{A+3A}{2}\cos\frac{A-3A}{2} \]

Prove that cos x cos(x/2) − cos 3x cos(9x/2) = sin 7x sin 8x Prove that: \[ \cos x \cos\frac{x}{2} – \cos 3x \cos\frac{9x}{2} = \sin 7x \sin 8x \] Solution Using the identity: \[ \cos A \cos B = \frac{1}{2} \left[ \cos(A+B)+\cos(A-B) \right] \] First term: \[ \cos x \cos\frac{x}{2} = \frac{1}{2} \left[ \cos\left(x+\frac{x}{2}\right)

Prove that sin(x/2) sin(7x/2) + sin(3x/2) sin(11x/2) = sin 2x sin 5x Prove that: \[ \sin\frac{x}{2}\sin\frac{7x}{2} + \sin\frac{3x}{2}\sin\frac{11x}{2} = \sin2x\sin5x \] Solution Using the identity: \[ \sin A \sin B = \frac{1}{2} \left[ \cos(A-B)-\cos(A+B) \right] \] First term: \[ \sin\frac{x}{2}\sin\frac{7x}{2} = \frac{1}{2} \left[ \cos\left(\frac{x}{2}-\frac{7x}{2}\right) – \cos\left(\frac{x}{2}+\frac{7x}{2}\right) \right] \] \[ = \frac{1}{2} \left[ \cos(-3x)-\cos4x \right] \]

Prove that cos 20° cos 100° + cos 100° cos 140° − cos 140° cos 20° = −3/4 Prove that: \[ \cos20^\circ\cos100^\circ + \cos100^\circ\cos140^\circ – \cos140^\circ\cos20^\circ = -\frac{3}{4} \] Solution Use the identity: \[ \cos A \cos B = \frac{1}{2} \left[ \cos(A+B)+\cos(A-B) \right] \] First term: \[ \cos20^\circ\cos100^\circ = \frac{1}{2} \left[ \cos120^\circ+\cos(-80^\circ) \right] \] \[