Class 11: Values of Trigonometric Functions at Multiples and Submultiples of an Angle

$\displaystyle \textbf{Question 1: }\text{Prove that }\sin^2\frac{2\pi}{5}-\sin^2\frac{\pi}{3}=\frac{\sqrt5-1}{8}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\sin^272^\circ-\sin^260^\circ$
$\displaystyle =\sin^2(90^\circ-18^\circ)-\left(\frac{\sqrt3}{2}\right)^2$
$\displaystyle =\cos^218^\circ-\frac{3}{4}$
$\displaystyle \text{Using }\cos18^\circ=\frac{\sqrt{10+2\sqrt5}}{4},$
$\displaystyle \text{LHS}=\left(\frac{\sqrt{10+2\sqrt5}}{4}\right)^2-\frac{3}{4}$
$\displaystyle =\frac{10+2\sqrt5}{16}-\frac{12}{16}$
$\displaystyle =\frac{2\sqrt5-2}{16}$
$\displaystyle =\frac{\sqrt5-1}{8}$
$\displaystyle =\text{RHS}$
$\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Prove that }\sin^224^\circ-\sin^26^\circ=\frac{\sqrt5-1}{8}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\sin^224^\circ-\sin^26^\circ$
$\displaystyle \text{Using }\sin(A+B)\sin(A-B)=\sin^2A-\sin^2B,$
$\displaystyle \text{LHS}=\sin(24^\circ+6^\circ)\sin(24^\circ-6^\circ)$
$\displaystyle =\sin30^\circ\sin18^\circ$
$\displaystyle \text{Using }\sin18^\circ=\frac{\sqrt5-1}{4},$
$\displaystyle \text{LHS}=\frac{1}{2}\times\frac{\sqrt5-1}{4}$
$\displaystyle =\frac{\sqrt5-1}{8}$
$\displaystyle =\text{RHS}$
$\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Prove that }\sin^242^\circ-\cos^278^\circ=\frac{\sqrt5+1}{8}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\sin^242^\circ-\cos^278^\circ$
$\displaystyle =\sin^2(90^\circ-48^\circ)-\cos^2(90^\circ-12^\circ)$
$\displaystyle =\cos^248^\circ-\sin^212^\circ$
$\displaystyle \text{Using }\cos(A+B)\cos(A-B)=\cos^2A-\sin^2B,$
$\displaystyle \text{LHS}=\cos(48^\circ+12^\circ)\cos(48^\circ-12^\circ)$
$\displaystyle =\cos60^\circ\cos36^\circ$
$\displaystyle \text{Using }\cos36^\circ=\frac{\sqrt5+1}{4},$
$\displaystyle \text{LHS}=\frac{1}{2}\times\frac{\sqrt5+1}{4}$
$\displaystyle =\frac{\sqrt5+1}{8}$
$\displaystyle =\text{RHS}$
$\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Prove that }\cos78^\circ\cos42^\circ\cos36^\circ=\frac{1}{8}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\cos78^\circ\cos42^\circ\cos36^\circ$
$\displaystyle =\frac{1}{2}\left[2\cos78^\circ\cos42^\circ\right]\cos36^\circ$
$\displaystyle =\frac{1}{2}\left[\cos(78^\circ+42^\circ)+\cos(78^\circ-42^\circ)\right]\cos36^\circ$
$\displaystyle =\frac{1}{2}\left[\cos120^\circ+\cos36^\circ\right]\cos36^\circ$
$\displaystyle =\frac{1}{2}\left[-\frac{1}{2}+\frac{\sqrt5+1}{4}\right]\left(\frac{\sqrt5+1}{4}\right)$
$\displaystyle =\frac{1}{2}\left(\frac{\sqrt5-1}{4}\right)\left(\frac{\sqrt5+1}{4}\right)$
$\displaystyle =\frac{1}{2}\times\frac{5-1}{16}$
$\displaystyle =\frac{1}{8}$
$\displaystyle =\text{RHS}$
$\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Prove that }\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{7\pi}{15}=\frac{1}{16}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{7\pi}{15}$
$\displaystyle =\frac{1}{2\sin\frac{\pi}{15}}\left(2\sin\frac{\pi}{15}\cos\frac{\pi}{15}\right)\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{7\pi}{15}$
$\displaystyle =\frac{1}{4\sin\frac{\pi}{15}}\left(2\sin\frac{2\pi}{15}\cos\frac{2\pi}{15}\right)\cos\frac{4\pi}{15}\cos\frac{7\pi}{15}$
$\displaystyle =\frac{1}{8\sin\frac{\pi}{15}}\left(2\sin\frac{4\pi}{15}\cos\frac{4\pi}{15}\right)\cos\frac{7\pi}{15}$
$\displaystyle =\frac{1}{16\sin\frac{\pi}{15}}\left(2\sin\frac{8\pi}{15}\cos\frac{7\pi}{15}\right)$
$\displaystyle =\frac{1}{16\sin\frac{\pi}{15}}\left[\sin\left(\frac{8\pi}{15}+\frac{7\pi}{15}\right)+\sin\left(\frac{8\pi}{15}-\frac{7\pi}{15}\right)\right]$
$\displaystyle =\frac{1}{16\sin\frac{\pi}{15}}\left(\sin\pi+\sin\frac{\pi}{15}\right)$
$\displaystyle =\frac{1}{16\sin\frac{\pi}{15}}\sin\frac{\pi}{15}$
$\displaystyle =\frac{1}{16}$
$\displaystyle =\text{RHS}$
$\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Prove that }\cos6^\circ\cos42^\circ\cos66^\circ\cos78^\circ=\frac{1}{16}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\cos6^\circ\cos42^\circ\cos66^\circ\cos78^\circ$
$\displaystyle =\frac{1}{4}\left(2\cos6^\circ\cos66^\circ\right)\left(2\cos42^\circ\cos78^\circ\right)$
$\displaystyle =\frac{1}{4}\left(\cos72^\circ+\cos60^\circ\right)\left(\cos120^\circ+\cos36^\circ\right)$
$\displaystyle =\frac{1}{4}\left(\sin18^\circ+\frac{1}{2}\right)\left(-\frac{1}{2}+\frac{\sqrt5+1}{4}\right)$
$\displaystyle =\frac{1}{4}\left(\frac{\sqrt5-1}{4}+\frac{2}{4}\right)\left(\frac{\sqrt5+1}{4}-\frac{2}{4}\right)$
$\displaystyle =\frac{1}{4}\left(\frac{\sqrt5+1}{4}\right)\left(\frac{\sqrt5-1}{4}\right)$
$\displaystyle =\frac{1}{4}\times\frac{5-1}{16}$
$\displaystyle =\frac{1}{16}$
$\displaystyle =\text{RHS}$
$\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Prove that }\sin6^\circ\sin42^\circ\sin66^\circ\sin78^\circ=\frac{1}{16}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\sin6^\circ\sin42^\circ\sin66^\circ\sin78^\circ$
$\displaystyle =\frac{1}{4}\left(2\sin6^\circ\sin66^\circ\right)\left(2\sin42^\circ\sin78^\circ\right)$
$\displaystyle =\frac{1}{4}\left(\cos60^\circ-\cos72^\circ\right)\left(\cos36^\circ-\cos120^\circ\right)$
$\displaystyle =\frac{1}{4}\left(\frac{1}{2}-\sin18^\circ\right)\left(\frac{\sqrt5+1}{4}+\frac{1}{2}\right)$
$\displaystyle =\frac{1}{4}\left(\frac{1}{2}-\frac{\sqrt5-1}{4}\right)\left(\frac{\sqrt5+1}{4}+\frac{2}{4}\right)$
$\displaystyle =\frac{1}{4}\left(\frac{3-\sqrt5}{4}\right)\left(\frac{3+\sqrt5}{4}\right)$
$\displaystyle =\frac{1}{4}\times\frac{9-5}{16}$
$\displaystyle =\frac{1}{16}$
$\displaystyle =\text{RHS}$
$\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Prove that }\cos36^\circ\cos42^\circ\cos60^\circ\cos78^\circ=\frac{1}{16}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\cos36^\circ\cos42^\circ\cos60^\circ\cos78^\circ$
$\displaystyle =\frac{1}{2}\cos36^\circ\cos60^\circ\left(2\cos42^\circ\cos78^\circ\right)$
$\displaystyle =\frac{1}{2}\left(\frac{\sqrt5+1}{4}\right)\times\frac{1}{2}\left(\cos120^\circ+\cos36^\circ\right)$
$\displaystyle =\frac{\sqrt5+1}{16}\left(-\frac{1}{2}+\frac{\sqrt5+1}{4}\right)$
$\displaystyle =\frac{\sqrt5+1}{16}\times\frac{\sqrt5-1}{4}$
$\displaystyle =\frac{(\sqrt5+1)(\sqrt5-1)}{64}$
$\displaystyle =\frac{5-1}{64}$
$\displaystyle =\frac{1}{16}$
$\displaystyle =\text{RHS}$
$\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{Prove that }\sin\frac{\pi}{5}\sin\frac{2\pi}{5}\sin\frac{3\pi}{5}\sin\frac{4\pi}{5}=\frac{5}{16}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\sin36^\circ\sin72^\circ\sin108^\circ\sin144^\circ$
$\displaystyle =\sin36^\circ\sin72^\circ\sin(180^\circ-72^\circ)\sin(180^\circ-36^\circ)$
$\displaystyle =\sin^236^\circ\sin^272^\circ$
$\displaystyle =\left(\sin36^\circ\sin72^\circ\right)^2$
$\displaystyle =\left(\sin36^\circ\cos18^\circ\right)^2$
$\displaystyle \text{Using }\sin36^\circ=\frac{\sqrt{10-2\sqrt5}}{4}\text{ and }\cos18^\circ=\frac{\sqrt{10+2\sqrt5}}{4},$
$\displaystyle \text{LHS}=\left(\frac{\sqrt{10-2\sqrt5}}{4}\times\frac{\sqrt{10+2\sqrt5}}{4}\right)^2$
$\displaystyle =\left(\frac{\sqrt{(10-2\sqrt5)(10+2\sqrt5)}}{16}\right)^2$
$\displaystyle =\left(\frac{\sqrt{100-20}}{16}\right)^2$
$\displaystyle =\left(\frac{\sqrt{80}}{16}\right)^2$
$\displaystyle =\frac{80}{256}$
$\displaystyle =\frac{5}{16}$
$\displaystyle =\text{RHS}$
$\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{Prove that }$
$\displaystyle \cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{3\pi}{15}\cos\frac{4\pi}{15}\cos\frac{5\pi}{15}\cos\frac{6\pi}{15}\cos\frac{7\pi}{15}=\frac{1}{128}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{3\pi}{15}\cos\frac{4\pi}{15}\cos\frac{5\pi}{15}\cos\frac{6\pi}{15}\cos\frac{7\pi}{15}$
$\displaystyle =\frac{1}{2}\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{3\pi}{15}\cos\frac{4\pi}{15}\cos\frac{6\pi}{15}\cos\frac{7\pi}{15}$
$\displaystyle =\frac{1}{4\sin\frac{\pi}{15}}\left(2\sin\frac{\pi}{15}\cos\frac{\pi}{15}\right)\cos\frac{2\pi}{15}\cos\frac{3\pi}{15}\cos\frac{4\pi}{15}\cos\frac{6\pi}{15}\cos\frac{7\pi}{15}$
$\displaystyle =\frac{1}{8\sin\frac{\pi}{15}}\left(2\sin\frac{2\pi}{15}\cos\frac{2\pi}{15}\right)\cos\frac{3\pi}{15}\cos\frac{4\pi}{15}\cos\frac{6\pi}{15}\cos\frac{7\pi}{15}$
$\displaystyle =\frac{1}{16\sin\frac{\pi}{15}}\left(2\sin\frac{4\pi}{15}\cos\frac{4\pi}{15}\right)\cos\frac{3\pi}{15}\cos\frac{6\pi}{15}\cos\frac{7\pi}{15}$
$\displaystyle =\frac{1}{32\sin\frac{\pi}{15}\sin\frac{3\pi}{15}}\left(2\sin\frac{3\pi}{15}\cos\frac{3\pi}{15}\right)\sin\frac{8\pi}{15}\cos\frac{6\pi}{15}\cos\frac{7\pi}{15}$
$\displaystyle =\frac{1}{64\sin\frac{\pi}{15}\sin\frac{3\pi}{15}}\left(2\sin\frac{6\pi}{15}\cos\frac{6\pi}{15}\right)\sin\frac{8\pi}{15}\cos\frac{7\pi}{15}$
$\displaystyle =\frac{1}{64\sin\frac{\pi}{15}\sin\frac{3\pi}{15}}\sin\frac{12\pi}{15}\sin\frac{8\pi}{15}\cos\frac{7\pi}{15}$
$\displaystyle =\frac{1}{64\sin\frac{\pi}{15}\sin\frac{3\pi}{15}}\sin\frac{3\pi}{15}\sin\frac{7\pi}{15}\cos\frac{7\pi}{15}$
$\displaystyle =\frac{1}{128\sin\frac{\pi}{15}}\left(2\sin\frac{7\pi}{15}\cos\frac{7\pi}{15}\right)$
$\displaystyle =\frac{1}{128\sin\frac{\pi}{15}}\sin\frac{14\pi}{15}$
$\displaystyle =\frac{1}{128\sin\frac{\pi}{15}}\sin\left(\pi-\frac{\pi}{15}\right)$
$\displaystyle =\frac{1}{128\sin\frac{\pi}{15}}\sin\frac{\pi}{15}$
$\displaystyle =\frac{1}{128}$
$\displaystyle =\text{RHS}$
$\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}$
$\displaystyle \\$