Prove that sin 10° sin 50° sin 60° sin 70° = √3/16 Prove that: \( \sin10^\circ \sin50^\circ \sin60^\circ \sin70^\circ = \frac{\sqrt3}{16} \) Solution: \[ \sin10^\circ \sin50^\circ \sin60^\circ \sin70^\circ \] Using identity, \[ \sin3\theta = 4\sin\theta \sin(60^\circ+\theta)\sin(60^\circ-\theta) \] Putting \( \theta=10^\circ \), \[ \sin30^\circ = 4\sin10^\circ \sin50^\circ \sin70^\circ \] \[ \frac12 = 4\sin10^\circ \sin50^\circ \sin70^\circ \] […]

Prove that sin 20° sin 40° sin 60° sin 80° = 3/16 Prove that: \( \sin20^\circ \sin40^\circ \sin60^\circ \sin80^\circ = \frac{3}{16} \) Solution: \[ \sin20^\circ \sin40^\circ \sin60^\circ \sin80^\circ \] Using identity, \[ \sin3\theta = 4\sin\theta \sin(60^\circ+\theta)\sin(60^\circ-\theta) \] Putting \( \theta=20^\circ \), \[ \sin60^\circ = 4\sin20^\circ \sin40^\circ \sin80^\circ \] \[ \frac{\sqrt3}{2} = 4\sin20^\circ \sin40^\circ \sin80^\circ \]

Prove that tan 20° tan 30° tan 40° tan 80° = 1 Prove that: \( \tan20^\circ \tan30^\circ \tan40^\circ \tan80^\circ = 1 \) Solution: \[ \tan20^\circ \tan30^\circ \tan40^\circ \tan80^\circ \] Using identity, \[ \tan\theta \tan(60^\circ-\theta)\tan(60^\circ+\theta)=\tan3\theta \] Putting \( \theta=20^\circ \), \[ \tan20^\circ \tan40^\circ \tan80^\circ = \tan60^\circ = \sqrt3 \] Therefore, \[ \tan20^\circ \tan30^\circ \tan40^\circ \tan80^\circ =

Prove that tan 20° tan 40° tan 60° tan 80° = 3 Prove that: \( \tan20^\circ \tan40^\circ \tan60^\circ \tan80^\circ = 3 \) Solution: \[ \tan20^\circ \tan40^\circ \tan60^\circ \tan80^\circ \] Using identity, \[ \tan\theta \tan(60^\circ-\theta)\tan(60^\circ+\theta)=\tan3\theta \] Putting \( \theta=20^\circ \), \[ \tan20^\circ \tan40^\circ \tan80^\circ = \tan60^\circ = \sqrt3 \] Therefore, \[ \tan20^\circ \tan40^\circ \tan60^\circ \tan80^\circ =

Prove that cos 20° cos 40° cos 80° = 1/8 Prove that \( \cos20^\circ\cos40^\circ\cos80^\circ=\frac18 \) Solution Using identity: \(2\cos A\cos B=\cos(A+B)+\cos(A-B)\) \[ \begin{aligned} &\cos20^\circ\cos40^\circ\cos80^\circ\\[4pt] &=\frac12(\cos60^\circ+\cos20^\circ)\cos80^\circ\\[4pt] &=\frac12\left(\frac12+\cos20^\circ\right)\cos80^\circ\\[4pt] &=\frac14\cos80^\circ+\frac12\cos20^\circ\cos80^\circ \end{aligned} \] Using identity: \(2\cos A\cos B=\cos(A+B)+\cos(A-B)\) \[ \begin{aligned} &=\frac14\cos80^\circ+\frac14(\cos100^\circ+\cos60^\circ)\\[4pt] &=\frac14\cos80^\circ+\frac14(-\cos80^\circ+\frac12)\\[4pt] &=\frac18 \end{aligned} \] “` Next Question / Full Exercise

Prove that sin 20° sin 40° sin 80° = √3/8 Prove that \( \sin20^\circ\sin40^\circ\sin80^\circ=\frac{\sqrt3}{8} \) Solution Using identity: \(2\sin A\sin B=\cos(A-B)-\cos(A+B)\) \[ \begin{aligned} &\sin20^\circ\sin40^\circ\sin80^\circ\\[4pt] &=\frac12(\cos20^\circ-\cos60^\circ)\sin80^\circ\\[4pt] &=\frac12\left(\cos20^\circ-\frac12\right)\sin80^\circ\\[4pt] &=\frac12\cos20^\circ\sin80^\circ-\frac14\sin80^\circ \end{aligned} \] Using identity: \(2\sin A\cos B=\sin(A+B)+\sin(A-B)\) \[ \begin{aligned} &=\frac14(\sin100^\circ+\sin60^\circ)-\frac14\sin80^\circ\\[4pt] &=\frac14(\cos10^\circ+\frac{\sqrt3}{2})-\frac14\cos10^\circ\\[4pt] &=\frac{\sqrt3}{8} \end{aligned} \] “` Next Question / Full Exercise

Prove that cos 40° cos 80° cos 160° = -1/8 Prove that \( \cos40^\circ\cos80^\circ\cos160^\circ=-\frac18 \) Solution \[ \begin{aligned} &\cos40^\circ\cos80^\circ\cos160^\circ\\[4pt] &=-\cos40^\circ\cos80^\circ\cos20^\circ \end{aligned} \] Using identity: \(2\cos A\cos B=\cos(A+B)+\cos(A-B)\) \[ \begin{aligned} &=-\frac12(\cos100^\circ+\cos60^\circ)\cos20^\circ\\[4pt] &=-\frac12\left(-\sin10^\circ+\frac12\right)\cos20^\circ \end{aligned} \] Using identity: \(2\sin A\cos B=\sin(A+B)+\sin(A-B)\) \[ \begin{aligned} &=-\frac14\cos20^\circ+\frac14(\sin30^\circ-\sin10^\circ)\\[4pt] &=-\frac14\cos20^\circ+\frac14\left(\frac12-\sin10^\circ\right)\\[4pt] &=-\frac18 \end{aligned} \] “` Next Question / Full Exercise

Prove that cos 10° cos 30° cos 50° cos 70° = 3/16 Prove that \( \cos10^\circ\cos30^\circ\cos50^\circ\cos70^\circ=\frac{3}{16} \) Solution \[ \begin{aligned} &\cos10^\circ\cos30^\circ\cos50^\circ\cos70^\circ\\[4pt] &=\frac{\sqrt3}{2}\cos10^\circ\cos50^\circ\sin20^\circ \end{aligned} \] Using identity: \(2\cos A\cos B=\cos(A+B)+\cos(A-B)\) \[ \begin{aligned} &=\frac{\sqrt3}{4}(\cos60^\circ+\cos40^\circ)\sin20^\circ\\[4pt] &=\frac{\sqrt3}{4}\left(\frac12+\cos40^\circ\right)\sin20^\circ\\[4pt] &=\frac{\sqrt3}{8}\sin20^\circ+\frac{\sqrt3}{4}\cos40^\circ\sin20^\circ \end{aligned} \] Using identity: \(2\sin A\cos B=\sin(A+B)-\sin(B-A)\) \[ \begin{aligned} &=\frac{\sqrt3}{8}\sin20^\circ+\frac{\sqrt3}{8}(\sin60^\circ-\sin20^\circ)\\[4pt] &=\frac{\sqrt3}{8}\times\frac{\sqrt3}{2}\\[4pt] &=\frac{3}{16} \end{aligned} \] “` Next Question / Full Exercise

Prove that 4 cos x cos(π/3 + x) cos(π/3 − x) = cos 3x Prove that \(4\cos x\cos\left(\frac{\pi}{3}+x\right)\cos\left(\frac{\pi}{3}-x\right)=\cos3x\) Solution \[ \begin{aligned} &4\cos x\cos\left(\frac{\pi}{3}+x\right)\cos\left(\frac{\pi}{3}-x\right)\\[4pt] &=2\cos x\left[2\cos\left(\frac{\pi}{3}+x\right)\cos\left(\frac{\pi}{3}-x\right)\right]\\[4pt] &=2\cos x\left[\cos\frac{2\pi}{3}+\cos2x\right]\\[4pt] &=2\cos x\left[-\frac12+\cos2x\right]\\[4pt] &=-\cos x+2\cos x\cos2x\\[4pt] &=-\cos x+\cos3x+\cos x\\[4pt] &=\cos3x \end{aligned} \] Next Question / Full Exercise

Show that sin 25° cos 115° = 1/2(sin 140° − 1) | Trigonometric Identities Show that \( \sin25^\circ\cos115^\circ=\frac{1}{2}(\sin140^\circ-1) \) Solution Using the identity: \[ 2\sin A\cos B=\sin(A+B)+\sin(A-B) \] \[ 2\sin25^\circ\cos115^\circ \] \[ = \sin(25^\circ+115^\circ)+\sin(25^\circ-115^\circ) \] \[ = \sin140^\circ+\sin(-90^\circ) \] \[ = \sin140^\circ-1 \] Dividing both sides by \(2\), \[ \sin25^\circ\cos115^\circ = \frac{1}{2}(\sin140^\circ-1) \] Hence Proved