Question

Show that

i) $\sin \frac{\pi }{6}\cos 0+\sin \frac{\pi }{4}\cos \frac{\pi }{4}+\sin \frac{\pi }{3}\cos \frac{\pi }{6}=\frac{7}{4}$

ii) ${\sin }^2\frac{\pi }{6}+{\cos }^2\frac{\pi }{3}-{\tan }^2\frac{\pi }{4}=-\frac{1}{2}$

Answer 1

Part (1):-

$LHS.$

$\sin \frac{\pi }{6}\cos 0^0+\sin \frac{\pi }{4}\cos \frac{\pi }{4}+\sin \frac{\pi }{3}\cos \frac{\pi }{6}$

$=\frac{1}{2}\times 1+\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}$

$=\frac{1}{2}+\frac{1}{2}+\frac{3}{4}$

$=\frac{2+2+3}{4}$

$=\frac{7}{4}RHS.$

Part (2):-

$LHS.$

${\sin }^2\frac{\pi }{6}+{\cos }^2\frac{\pi }{3}-{\tan }^2\frac{\pi }{4}$

$={\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2-{\left(1\right)}^2$

$=\frac{1}{4}+\frac{1}{4}-1$

$=\frac{1+1-4}{4}$

$=\frac{-2}{4}$

$=\frac{-1}{2}$

$RHS.$