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1) \(2sin^2\theta-1=sin^4\theta-cos^4\theta\)

2) \(\frac{cos^2\theta-csc^2\theta+sin^2\theta}{sin\theta sec\theta}=-cot^3\theta\)

AdamTaurus  Apr 1, 2018
 #1
avatar+86944 
+2

1) 

 

2sin^2A - 1  =   sin^4A - cos^4A

Working with the right side we can factor this as

 

(sin2^A + cos^2A) (sin^2A - cos^2A)  =

(1) ( sin^2 A - cos^2A)  =

sin^2A  -  (1  - sin^2A) =

sin^2A  - 1 + sin^2 A  =

2sin^2A  - 1        which is = to the left side

 

 

2) 

 

[ cos^2A - csc^2A + sin^2A]

     _______________________   =  - cot^3A

          sinAsec A

 

Obviously...the left side is the side that needs to be simplfied

Note......  sin^2A +cos^A  = 1

 

  [  1  - csc^2]

______________

   sinA (1/cos A]

 

Note

cot^2A  + 1  = csc^2A   ....so...

1 - csc^2A = - cot^2A

 

-cot^2A  * cosA

____________

   sinA

 

-cot^2A  *  cosA

                 ______

                  sin A

 

-cot^2A * cot A

 

-cot^3A       which is the right side

 

 

cool cool cool

CPhill  Apr 1, 2018

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