How do you find the value of #sec((5pi)/4)#?

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Answer 1 · 2018-04-06T09:19:40

#color(blue)(-sqrt(2))#

Identities:

#color(red)bb(secx=1/cosx) \ \ \ \ \[1]#

#color(red)bb(cos(A+B)=cosAcosB-sinAsinB) \ \ \ \[2]#

#cos((5pi)/4)=cos(pi+pi/4)#

Using #[2]#

#cos(pi+pi/4)=cos(pi)cos(pi/4)-sin(pi)sin(pi/4)#

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =(-1)(sqrt(2)/2)-(0)((sqrt(2))/2)#

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-sqrt(2)/2#

Using #[1]#

#1/cosx\ \ \ \ \ \ \ \ \ \ \ \ =-1/(sqrt(2)/2)=-sqrt(2)#

#:.#

#color(blue)(sec((5pi)/4)=-sqrt(2))#