Prove that cos6° cos42° cos66° cos78° = 1/16 Prove that: \[ \cos6^\circ\cos42^\circ\cos66^\circ\cos78^\circ = \frac1{16} \] Solution Using \[ \cos78^\circ=\sin12^\circ \] \[ \cos66^\circ=\sin24^\circ \] therefore, \[ \cos6^\circ\cos42^\circ\cos66^\circ\cos78^\circ = \cos6^\circ\cos42^\circ\sin24^\circ\sin12^\circ \] Now use \[ 2\sin A\cos B = \sin(A+B)+\sin(A-B) \] For \[ A=24^\circ,\qquad B=42^\circ \] \[ 2\sin24^\circ\cos42^\circ = \sin66^\circ+\sin(-18^\circ) \] \[ = \sin66^\circ-\sin18^\circ \] \[ = \cos24^\circ-\sin18^\circ […]

Prove that cos(π/15) cos(2π/15) cos(4π/15) cos(7π/15) = 1/16 Prove that: \[ \cos\frac{\pi}{15} \cos\frac{2\pi}{15} \cos\frac{4\pi}{15} \cos\frac{7\pi}{15} = \frac1{16} \] Solution Using the identity \[ 2\sin A\cos A=\sin2A \] we successively write: \[ \sin\frac{2\pi}{15} = 2\sin\frac{\pi}{15}\cos\frac{\pi}{15} \] \[ \sin\frac{4\pi}{15} = 2\sin\frac{2\pi}{15}\cos\frac{2\pi}{15} \] \[ \sin\frac{8\pi}{15} = 2\sin\frac{4\pi}{15}\cos\frac{4\pi}{15} \] \[ \sin\frac{16\pi}{15} = 2\sin\frac{8\pi}{15}\cos\frac{8\pi}{15} \] Multiply all these equations: \[

Prove that cos78° cos42° cos36° = 1/8 Prove that: \[ \cos78^\circ\cos42^\circ\cos36^\circ=\frac18 \] Solution Using \[ \cos78^\circ=\sin12^\circ \] therefore, \[ \cos78^\circ\cos42^\circ\cos36^\circ = \sin12^\circ\cos42^\circ\cos36^\circ \] Now use \[ 2\sin A\cos B = \sin(A+B)+\sin(A-B) \] with \[ A=12^\circ,\qquad B=42^\circ \] Then \[ 2\sin12^\circ\cos42^\circ = \sin54^\circ+\sin(-30^\circ) \] \[ = \sin54^\circ-\sin30^\circ \] \[ = \cos36^\circ-\frac12 \] Hence, \[ \sin12^\circ\cos42^\circ =

Prove that sin²42° − cos²78° = (√5 + 1)/8 Prove that: \[ \sin^2 42^\circ-\cos^2 78^\circ = \frac{\sqrt5+1}{8} \] Solution Using \[ \cos(90^\circ-\theta)=\sin\theta \] we get \[ \cos78^\circ=\sin12^\circ \] Therefore, \[ \sin^2 42^\circ-\cos^2 78^\circ = \sin^2 42^\circ-\sin^2 12^\circ \] Using the identity \[ \sin^2A-\sin^2B = \sin(A+B)\sin(A-B) \] Let \[ A=42^\circ,\qquad B=12^\circ \] Then \[ A+B=54^\circ \]

Prove that sin²24° − sin²6° = (√5 − 1)/8 Prove that: \[ \sin^2 24^\circ-\sin^2 6^\circ = \frac{\sqrt5-1}{8} \] Solution Using the identity \[ \sin^2A-\sin^2B = \sin(A+B)\sin(A-B) \] Let \[ A=24^\circ,\qquad B=6^\circ \] Then \[ A+B=30^\circ \] \[ A-B=18^\circ \] Therefore, \[ \sin^2 24^\circ-\sin^2 6^\circ = \sin30^\circ\sin18^\circ \] \[ = \frac12\sin18^\circ \] Now use the standard

Value of Trigonometric Functions at Multiples and Submultiples of an angle – Exercise 9.3 Solutions Prove that: sin² 2π/5 – sin² π/3 = (√5 – 1)/8  Watch Solution Prove that: sin² 24° – sin² 6° = (√5 – 1)/8 Watch Solution Prove that: sin² 42° – cos² 78° = (√5 + 1)/8 Watch Solution Prove that:

Prove that |cosx cos(π/3 − x) cos(π/3 + x)| ≤ 1/4 Prove that: \[ \left| \cos x\cos\left(\frac{\pi}{3}-x\right) \cos\left(\frac{\pi}{3}+x\right) \right| \le \frac14 \] for all values of \(x\). Solution Consider \[ \cos\left(\frac{\pi}{3}-x\right) \cos\left(\frac{\pi}{3}+x\right) \] Using the identity \[ \cos(A-B)\cos(A+B) = \cos^2A-\sin^2B \] with \[ A=\frac{\pi}{3}, \qquad B=x \] we get \[ \cos\left(\frac{\pi}{3}-x\right) \cos\left(\frac{\pi}{3}+x\right) = \cos^2\frac{\pi}{3}-\sin^2x \]

Prove that sin²(2π/5) − sin²(π/10) = (√5 − 1)/8 Prove that: \[ \sin^2\frac{2\pi}{5} – \sin^2\frac{\pi}{10} = \frac{\sqrt5-1}{8} \] Solution Using the identity \[ \sin^2A-\sin^2B = \sin(A+B)\sin(A-B) \] Let \[ A=\frac{2\pi}{5}, \qquad B=\frac{\pi}{10} \] Then \[ A+B = \frac{2\pi}{5}+\frac{\pi}{10} = \frac{5\pi}{10} = \frac{\pi}{2} \] \[ A-B = \frac{2\pi}{5}-\frac{\pi}{10} = \frac{3\pi}{10} \] Therefore, \[ \sin^2\frac{2\pi}{5} – \sin^2\frac{\pi}{10}

Prove that |sinx sin(π/3 − x) sin(π/3 + x)| ≤ 1/4 Prove that: \[ \left| \sin x\sin\left(\frac{\pi}{3}-x\right) \sin\left(\frac{\pi}{3}+x\right) \right| \le \frac14 \] for all values of \(x\). Solution Consider \[ \sin\left(\frac{\pi}{3}-x\right) \sin\left(\frac{\pi}{3}+x\right) \] Using the identity \[ \sin(A-B)\sin(A+B) = \sin^2A-\sin^2B \] with \[ A=\frac{\pi}{3}, \qquad B=x \] we get \[ \sin\left(\frac{\pi}{3}-x\right) \sin\left(\frac{\pi}{3}+x\right) = \sin^2\frac{\pi}{3}-\sin^2x \]

Prove that sin³x + sin³(2π/3 + x) + sin³(4π/3 + x) = −3/4 sin3x Prove that: \[ \sin^3 x+\sin^3\left(\frac{2\pi}{3}+x\right) +\sin^3\left(\frac{4\pi}{3}+x\right) = -\frac{3}{4}\sin 3x \] Solution Using the identity \[ \sin^3\theta = \frac{3\sin\theta-\sin3\theta}{4} \] we get \[ \sin^3 x = \frac{3\sin x-\sin3x}{4} \] \[ \sin^3\left(\frac{2\pi}{3}+x\right) = \frac{ 3\sin\left(\frac{2\pi}{3}+x\right) – \sin\left(2\pi+3x\right) }{4} \] \[ \sin^3\left(\frac{4\pi}{3}+x\right) = \frac{