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avatar+315 

Pre-calculus/trig

 Apr 26, 2019
 #1
avatar+106535 
+3

We don't need the "squared identities," here...the double-angle identities serve better....

 

sin^2xcos^2x  = 

(sinx cosx)^2  =                                                                                

[  (1/2) sin 2x] ^2  =                                                         

 (1/4) (sin 2x)^2                       

 

Note, Cupcake.......     cos 4x  =   1 - 2[sin 2x]^2

                                    2[sin 2x ]^2  = 1 - cos(4x)

                                     (sin 2x)^2  =  [ 1 - cos(4x)] / 2   =  (1/2)[ 1 - cos (4x) ]

 

So we have

 

(1/4) (1/2)  (1 - cos(4x) ]    =

 

(1/8) [ 1 - cos (4x) ]  =

 

[ 1 - cos (4x) ]

____________

         8

 

 

cool cool cool                                                 

 Apr 26, 2019
 #2
avatar+315 
+1

Thank you!! smiley

Cupcake  Apr 26, 2019
 #3
avatar+107006 
+1

Thanks Chris,

That did not jump out at me. 

Melody  Apr 26, 2019
 #4
avatar+106535 
+2

Well....truth be told....it didn't jump out for me, either....LOL!!!!

 

 

cool cool cool

CPhill  Apr 26, 2019

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