We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
375
1
avatar+895 

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

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

 Apr 1, 2018
 #1
avatar+101761 
+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

 Apr 1, 2018

9 Online Users

avatar
avatar