Simplify (Sin^2((pi/2)-theta))/sec(-theta))?

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Answer 1 · 2018-02-07T16:48:19

# cos^3theta#.

Recall that, #sin(pi/2-theta)=costheta, and cos(-theta)=costheta#.

#:. sin^2(pi/2-theta)/sec(-theta)={sin(pi/2-theta)}^2-:{1/cos(-theta)}#,

#={costheta}^2-:(1/costheta)#,

#=cos^2theta*costheta#,

#=cos^3theta#.

Answer 2 · 2018-02-07T17:04:01

#cos^3(theta)#

#(sin^2(pi/2-theta))/(sec(-theta))#

Identity:

#color(red)bb(sin(A-B)=sinAcosB-cosAsinB#

#sin^2(pi/2-theta)=#

#(sin(pi/2)cos(theta)-cos(pi/2)sin(theta))^2#

#(1*cos(theta)-0*sin(theta))^2=cos^2(theta)#*

Identity:

#color(red)bb(secx=1/cosx#

#sec(-theta)=sec(theta)=1/cos(theta)#*

Since

#sec(theta)=1/cos(theta)# , and #cos(theta)# is an even function

.i.e. #cos(theta)=cos(-theta)#

#(cos^2(theta))/(1/cos(theta))=cos^3(theta)#