Educational

Write the value of (1 − 4 sin 10° sin 70°)/(2 sin 10°) Write the value of \( \dfrac{1-4\sin10^\circ\sin70^\circ}{2\sin10^\circ} \) Solution: Using identity, \[ 2\sin A\sin B = \cos(A-B)-\cos(A+B) \] \[ 2\sin10^\circ\sin70^\circ = \cos60^\circ-\cos80^\circ \] \[ = \frac12-\cos80^\circ \] Multiplying by 2, \[ 4\sin10^\circ\sin70^\circ = 1-2\cos80^\circ \] Therefore, \[ 1-4\sin10^\circ\sin70^\circ = 2\cos80^\circ \] Hence, \[ […]

If cos A = m cos B, then find cot((A + B)/2) cot((A − B)/2) If \( \cos A=m\cos B \), then write the value of \( \cot\frac{A+B}{2}\cot\frac{A-B}{2} \) Solution: Given, \[ \cos A=m\cos B \] \[ \frac{\cos A}{\cos B}=m \] Using identities, \[ \cos A+\cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} \] and \[ \cos A-\cos B

If sin A + sin B = α and cos A + cos B = β, then find tan((A + B)/2) If \( \sin A+\sin B=\alpha \) and \( \cos A+\cos B=\beta \), then write the value of \( \tan\frac{A+B}{2} \) Solution: Using identities, \[ \sin A+\sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} \] \[ \alpha = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}

Write the value of sin(π/12) sin(5π/12) Write the value of \( \sin\frac{\pi}{12}\sin\frac{5\pi}{12} \) Solution: Using identity, \[ \sin A\sin B = \frac12[\cos(A-B)-\cos(A+B)] \] \[ \sin\frac{\pi}{12}\sin\frac{5\pi}{12} \] \[ = \frac12 \left[ \cos\left(\frac{\pi}{12}-\frac{5\pi}{12}\right) – \cos\left(\frac{\pi}{12}+\frac{5\pi}{12}\right) \right] \] \[ = \frac12 \left[ \cos\left(-\frac{\pi}{3}\right) – \cos\left(\frac{\pi}{2}\right) \right] \] Using, \[ \cos(-\theta)=\cos\theta \] \[ = \frac12 \left[ \cos\frac{\pi}{3}-0 \right] \]

The value of 2 cos(π/13) cos(9π/13) + cos(3π/13) + cos(5π/13) The value of \( 2\cos\frac{\pi}{13}\cos\frac{9\pi}{13}+\cos\frac{3\pi}{13}+\cos\frac{5\pi}{13} \) is Solution: Using identity, \[ 2\cos A\cos B = \cos(A+B)+\cos(A-B) \] \[ 2\cos\frac{\pi}{13}\cos\frac{9\pi}{13} = \cos\frac{10\pi}{13}+\cos\frac{8\pi}{13} \] Therefore, \[ = \cos\frac{10\pi}{13} + \cos\frac{8\pi}{13} + \cos\frac{3\pi}{13} + \cos\frac{5\pi}{13} \] Using, \[ \cos(\pi-\theta)=-\cos\theta \] \[ \cos\frac{10\pi}{13} = -\cos\frac{3\pi}{13} \] and \[ \cos\frac{8\pi}{13}

If (cos α + cos β)² + (sin α + sin β)² = λ cos²((α − β)/2), find λ If \( (\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2=\lambda\cos^2\frac{\alpha-\beta}{2} \), find \( \lambda \) Solution: \[ (\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2 \] Expanding, \[ =\cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta \] \[ +\sin^2\alpha+\sin^2\beta+2\sin\alpha\sin\beta \] Using, \[ \sin^2\theta+\cos^2\theta=1 \] \[ =1+1+2(\cos\alpha\cos\beta+\sin\alpha\sin\beta) \] Using identity, \[ \cos(\alpha-\beta) = \cos\alpha\cos\beta+\sin\alpha\sin\beta \] \[ =2+2\cos(\alpha-\beta) \]

The value of sin 50° − sin 70° + sin 10° The value of \( \sin50^\circ-\sin70^\circ+\sin10^\circ \) is Solution: \[ =\sin50^\circ-\sin70^\circ+\sin10^\circ \] Grouping first two terms, \[ =(\sin50^\circ-\sin70^\circ)+\sin10^\circ \] Using identity, \[ \sin A-\sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2} \] \[ = 2\cos60^\circ\sin(-10^\circ)+\sin10^\circ \] \[ = 2\left(\frac12\right)(-\sin10^\circ)+\sin10^\circ \] \[ = -\sin10^\circ+\sin10^\circ \] \[ =0 \] \[ \boxed{0}

If 1 + cos 2x + cos 4x + cos 6x = k cos x cos 2x cos 3x, then find k If \( 1+\cos2x+\cos4x+\cos6x=k\cos x\cos2x\cos3x \), then \( k= \) Solution: \[ 1+\cos2x = 2\cos^2x \] Therefore, \[ 1+\cos2x+\cos4x+\cos6x \] \[ = 2\cos^2x+(\cos4x+\cos6x) \] Using identity, \[ \cos A+\cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} \] \[

If tan(A + B) = p and tan(A − B) = q, then find tan 2A If \( \tan(A+B)=p \) and \( \tan(A-B)=q \), then find \( \tan2A \) Solution: Since, \[ 2A=(A+B)+(A-B) \] Using identity, \[ \tan(x+y) = \frac{\tan x+\tan y}{1-\tan x\tan y} \] \[ \tan2A = \tan[(A+B)+(A-B)] \] \[ = \frac{\tan(A+B)+\tan(A-B)} {1-\tan(A+B)\tan(A-B)} \]

The value of (sin 70° + cos 40°)/(cos 70° + sin 40°) The value of \( \dfrac{\sin70^\circ+\cos40^\circ}{\cos70^\circ+\sin40^\circ} \) is Solution: Using, \[ \cos\theta=\sin(90^\circ-\theta) \] \[ \cos40^\circ=\sin50^\circ \] and \[ \cos70^\circ=\sin20^\circ \] Therefore, \[ = \frac{\sin70^\circ+\sin50^\circ} {\sin20^\circ+\sin40^\circ} \] Using identity, \[ \sin A+\sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} \] \[ = \frac{ 2\sin60^\circ\cos10^\circ } { 2\sin30^\circ\cos10^\circ } \]