# P3 Trig IdentitiesWatch

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#1
Hi
Sorry if these questions are easy, but I hate trig identities
How would I simplify the following?:
a) sec(1/2Pi - x)
b) cosec(Pi + x)
c) Why does sec(-x) = sec(x)? (Thats not the question; the answer in the back says its sec(x), but Im not sure why).
Thanks for any help
0
13 years ago
#2
a) sec(1/2Pi - x)
sec(0.5pi-x)=1/cos(0.5pi-x)=1/[cos0.5picosx+sin0.5pisinx]=1/(sinx)=cosecx. (Using the expression cos(A-B)=CosAcosB+sinAsinB)

b) cosec(Pi + x)
Similar to above but using the Sin(A+B) formula.

c) Why does sec(-x) = sec(x)? (Thats not the question; the answer in the back says its sec(x), but Im not sure why).
sec(-x)=1/cos(-x)=1/cosx=secx [Cos is an even function, not sure how you'd prove that though maybe the maclurin expansion would work, you aren't being asked that though].
0
13 years ago
#3
a) cos(A-B)=cosAcosB+sinAsinB

=>cos((Pi/2)-x)=cos(Pi/2)cosx+sin(Pi/2)sinx=sinx

=>(1/sec((Pi/2)-x))=(1/sinx)=cosecx

b) sin(A+B)=sinAcosB+sinBcosA

=>sin((Pi)+x)=sin(Pi)cosx+sinxco s(Pi)=cosx-sinx

=>cosec((Pi)+x)=(1/(cosx-sinx))

c) Let f(x)=secx=(1/cosx)

Since cosx is an even function =>f(-x)=f(x) it follows that sec(-x)=secx.

Newton.
0
#4
Thanks a lot for the help
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