Answer:
[tex]sin(x+\frac{\pi }{2}) = \frac{-4}{5}[/tex]
Step-by-step explanation:
Given [tex]cos x = \frac{-4}{5}[/tex]
given 'x' lies in [tex]\pi <x<\frac{3\pi }{2}[/tex]
we know that sin(A+B) = sin A cos B + cos A sin B
[tex]sin(x+\frac{\pi }{2} )[/tex] = [tex]sin x cos(\frac{\pi }{2} )+ cos x sin(\frac{\pi }{2} )[/tex] .........................(1)
we will use trigonometry formulas
[tex]cos(\frac{\pi }{2}) = 0[/tex]
[tex]sin(\frac{\pi }{2} )=1[/tex]
sin x is negative in third and fourth quadrant ( [tex]\pi <x<\frac{3\pi }{2}[/tex])
find sin x value
using trigonometry formulas [tex]sin^{2} x+cos^2 x=1[/tex]
[tex]sin^{2}x = 1 - cos^2 x[/tex]
= [tex]1-(\frac{-4}{5)})^2[/tex]
[tex]sin^2 x = 1 - \frac{16}{25}[/tex]
[tex]sin x = \sqrt{\frac{25-16}{25} }[/tex]
sin x = [tex]\sqrt{\frac{9}{25} }[/tex]
[tex]sin x = \frac{3}{5}[/tex]
in third and fourth quadrant is negative so sin x= [tex]\frac{-3}{5}[/tex]
now equation (1), we get solution
[tex]sin(x+\frac{\pi }{2}) = sin x ( 0 ) + cos x (1)[/tex]
= [tex]\frac{-4}{5}[/tex]
