Show that sin 50° cos 85° = (1 − √2 sin 35°) / (2√2) | Trigonometric Identities Show that \( \sin 50^\circ \cos 85^\circ = \dfrac{1-\sqrt{2}\sin35^\circ}{2\sqrt{2}} \) Solution Using the identity: \[ 2\sin A\cos B=\sin(A+B)+\sin(A-B) \] \[ 2\sin50^\circ\cos85^\circ \] \[ = \sin(50^\circ+85^\circ)+\sin(50^\circ-85^\circ) \] \[ = \sin135^\circ+\sin(-35^\circ) \] \[ = \frac{1}{\sqrt2}-\sin35^\circ \] Dividing both sides by […]

Prove that 2 sin(5π/12) cos(π/12) = (√3 + 1)/2 | Trigonometric Identities Prove that \(2\sin\frac{5\pi}{12}\cos\frac{\pi}{12}=\frac{\sqrt{3}+1}{2}\) Solution Using the identity: \[ 2\sin A\cos B=\sin(A+B)+\sin(A-B) \] \[ 2\sin\frac{5\pi}{12}\cos\frac{\pi}{12} \] \[ = \sin\left(\frac{5\pi}{12}+\frac{\pi}{12}\right) +\sin\left(\frac{5\pi}{12}-\frac{\pi}{12}\right) \] \[ = \sin\frac{6\pi}{12}+\sin\frac{4\pi}{12} \] \[ = \sin\frac{\pi}{2}+\sin\frac{\pi}{3} \] \[ = 1+\frac{\sqrt{3}}{2} \] \[ = \frac{2+\sqrt{3}}{2} \] Hence Proved \[ 2\sin\frac{5\pi}{12}\cos\frac{\pi}{12} =\frac{2+\sqrt{3}}{2} \] Next

Prove that 2 cos 5π/12 cos π/12 = 1/2 | Trigonometric Identities Prove that \(2\cos\frac{5\pi}{12}\cos\frac{\pi}{12}=\frac{1}{2}\) Solution Using the identity: \[ 2\cos A\cos B=\cos(A+B)+\cos(A-B) \] \[ 2\cos\frac{5\pi}{12}\cos\frac{\pi}{12} \] \[ = \cos\left(\frac{5\pi}{12}+\frac{\pi}{12}\right) +\cos\left(\frac{5\pi}{12}-\frac{\pi}{12}\right) \] \[ = \cos\frac{6\pi}{12}+\cos\frac{4\pi}{12} \] \[ = \cos\frac{\pi}{2}+\cos\frac{\pi}{3} \] \[ = 0+\frac{1}{2} \] \[ = \frac{1}{2} \] Hence Proved \[ 2\cos\frac{5\pi}{12}\cos\frac{\pi}{12}=\frac{1}{2} \] Next Question

Express Each as Sum or Difference of Sines and Cosines | Product to Sum Formula Class 11 Maths Express Each as the Sum or Difference of Sines and Cosines Using product-to-sum identities, express the following trigonometric products as sums or differences of sines and cosines. Product-to-Sum Identities Used \[ 2\sin A \cos B = \sin(A+B)

Prove that 2 sin 5π/12 sin π/12 = 1/2 | Trigonometric Identities Prove that \(2\sin\frac{5\pi}{12}\sin\frac{\pi}{12}=\frac{1}{2}\) We use the product-to-sum identity to prove the given trigonometric expression. Identity Used \[ 2\sin A\sin B=\cos(A-B)-\cos(A+B) \] Proof \[ 2\sin\frac{5\pi}{12}\sin\frac{\pi}{12} \] Applying the identity: \[ = \cos\left(\frac{5\pi}{12}-\frac{\pi}{12}\right) -\cos\left(\frac{5\pi}{12}+\frac{\pi}{12}\right) \] \[ = \cos\frac{4\pi}{12}-\cos\frac{6\pi}{12} \] \[ = \cos\frac{\pi}{3}-\cos\frac{\pi}{2} \] Using standard

If tan α = 1/(1 + 2−x) and tan β = 1/(1 + 2x), Find α + β If tan α = 1/(1 + 2−x) and tan β = 1/(1 + 2x), Find α + β Question: If \[ \tan\alpha=\frac{1}{1+2^{-x}} \] and \[ \tan\beta=\frac{1}{1+2^x} \] find the value of \[ \alpha+\beta \] lying in the

If sin α − sin β = a and cos α + cos β = b, Find cos(α + β) If sin α − sin β = a and cos α + cos β = b, Find cos(α + β) Question: If \[ \sin\alpha-\sin\beta=a \] and \[ \cos\alpha+\cos\beta=b \] find \[ \cos(\alpha+\beta) \] Solution Using,

If A + B = C, Find tan A tan B tan C If A + B = C, Find tan A tan B tan C Question: If \[ A+B=C \] find \[ \tan A\tan B\tan C \] Solution Using, \[ \tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\tan B} \] Since \[ A+B=C \] \[ \tan

If a = b cos(2π/3) = c cos(4π/3), Find ab + bc + ca If a = b cos(2π/3) = c cos(4π/3), Find ab + bc + ca Question: If \[ a=b\cos\frac{2\pi}{3}=c\cos\frac{4\pi}{3} \] find \[ ab+bc+ca \] Solution \[ \cos\frac{2\pi}{3}=-\frac12, \qquad \cos\frac{4\pi}{3}=-\frac12 \] Given, \[ a=-\frac{b}{2}=-\frac{c}{2} \] \[ b=-2a, \qquad c=-2a \] Now, \[ ab+bc+ca

If cos(x − y)/cos(x + y) = m/n, Find tan x tan y If cos(x − y)/cos(x + y) = m/n, Find tan x tan y Question: If \[ \frac{\cos(x-y)}{\cos(x+y)}=\frac{m}{n} \] find \[ \tan x\tan y \] Solution Using, \[ \cos(x-y)=\cos x\cos y+\sin x\sin y \] and \[ \cos(x+y)=\cos x\cos y-\sin x\sin y \] \[