Question

find the value of sin (npi+(-1)power n pi/3) where n is any integer​

Answer 1

Answer:

Step-by-step explanation:

sinx=sina  

then solution of trigonometry  

x=$x=n*\pi+-1^{n(a)}$

now here

$sin[{n\pi+(-1)^{n(\frac{\pi }{3 })] }=sin(\frac{\pi }{3} )=\frac{\sqrt{3} }{2}$

2nd method

$sin(n\pi +(-1)^{n(\frac{\pi }{3} )}  )$

put n=0

$= sin(\frac{\pi }{3} )=\frac{\sqrt{3} }{2}$

put n=1

$sin(\pi-\frac{\pi }{3}  )=sin (\frac{2\pi }{3} )$

since sin is in 2nd quadrant

so $sin(\frac{2\pi }{3} )=\frac{\sqrt{3} }{2}$

put

n=1

sin(-π - π/4)

=sin(-π/4)

=√3/2

so the answer will be √3/2

hope it help you