Question

Sin 8pi/3 cos 23pi/6 + cos 13pi/3 sin 35pi/6 = 1/2

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

sin$\frac{8\pi }{3}$. cos$\frac{23\pi }{6}$ + cos$\frac{13\pi }{3}$. sin$\frac{35\pi }{6}$ = $\frac{1}{2}$

Step-by-step explanation:

• We have to prove that, sin$\frac{8\pi }{3}$. cos$\frac{23\pi }{6}$ + cos$\frac{13\pi }{3}$. sin$\frac{35\pi }{6}$ = $\frac{1}{2}$

• sin$\frac{8\pi }{3}$ = sin ($3\pi -\frac{\pi }{3}$)

                   = sin$\frac{\pi }{3}$  

• cos$\frac{23\pi }{6}$ = cos$(4\pi - \frac{\pi }{6} )$

                   = cos$\frac{\pi }{6}$

• cos$\frac{13\pi }{3}$ = cos$(4\pi + \frac{\pi }{3})$

                   = cos$\frac{\pi }{3}$

• sin$\frac{35\pi }{6}$ = sin$(6\pi - \frac{\pi }{6})$

                  = - sin$\frac{\pi }{6}$

• sin$\frac{8\pi }{3}$. cos$\frac{23\pi }{6}$ + cos$\frac{13\pi }{3}$. sin$\frac{35\pi }{6}$

                       = sin$\frac{\pi }{3}$ cos$\frac{\pi }{6}$ - cos$\frac{\pi }{3}$ sin$\frac{\pi }{6}$

                      = sin ($\frac{\pi }{3}$ - $\frac{\pi }{6}$)      [as we know sin(A-B) = sinAcosB - cosAsinB]

                      = sin$\frac{\pi }{6}$

                      = $\frac{1}{2}$

∴sin$\frac{8\pi }{3}$. cos$\frac{23\pi }{6}$ + cos$\frac{13\pi }{3}$. sin$\frac{35\pi }{6}$ = $\frac{1}{2}$ [ proved]