How do you express #f(theta)=-cos^2(theta)-7sec^2(theta)-5sec^4theta# in terms of non-exponential trigonometric functions?

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Answer 1 · 2017-03-15T18:27:26

see below

Use the Double Argument Property

#cos 2 theta=2cos^2 theta -1#---> If we solve for #cos^2 theta# then we have #color(red)(cos^2 theta =1/2 (cos 2theta+1))#

#cos 4 theta=2cos^2 2theta -1#

#=2(cos 2 theta)^2-1#

#=2(2cos^2theta-1)^2-1#

#=2(4cos^4 theta-4 cos^2 theta +1)-1#

#=8cos^4 theta-8cos^2 theta +2-1#

#=8cos^4 theta-8cos^2 theta +1#--> If we put in #1/2 (cos 2theta+1)# for #cos^2 theta# we have

#=8cos^4 theta-8(1/2 (cos 2 theta+1)) +1#

#=8cos^4 theta-4 (cos 2 theta+1) +1#

#=8cos^4 theta-4 cos 2 theta-4 +1#

#=8cos^4 theta-4 cos 2 theta-3#---> Now solve for #cos^4 theta#

#color(red)(1/ 8 (cos 4 theta+4cos 2 theta+3)= cos^4 theta #

Therefore,

#f(theta)=-cos^2 theta-7 sec^2 theta-5 sec^4 theta#

#=-cos^2 theta -7/cos^2 theta - 5 / cos^4 theta#

#=-1/2 (cos 2theta+1)-7/(1/2 (cos 2theta+1)) - 5 /(1/ 8 (cos 4 theta+4cos 2 theta+3))#

#:.color(blue)(f(theta)=-1/2 (cos 2theta+1)-14/(cos 2theta+1)- 40/(cos 4 theta+4cos 2 theta+3)#