SOLUTION: If sin⁶β + cos⁶β = ¼, find (1/sin⁶β) + (1/cos⁶β)

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Question 1209544: If sin⁶β + cos⁶β = ¼,
find (1/sin⁶β) + (1/cos⁶β)
Answer by ikleyn(52747) About Me  (Show Source):
You can put this solution on YOUR website!
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If sin⁶β + cos⁶β = ¼,
find (1/sin⁶β) + (1/cos⁶β)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

Let  x = sin%28beta%29,  y = cos%28beta%29.


We have  x^2 + y^2 = 1    (Pythagorean identity)    (1).


Raise (1) to degree 3.  You will get

    x^6 + 3x^4*y^2 + 3x^2*y^4 + y^6 = 1,

    (x^6 + y^6) + 3x^2*y^2(x^2 + y^2) = 1.


Replace here  x^6 + y^6  by 1/4  (since it is given), and replace x^2 + y^2 by 1.  You will get

    1%2F4 + 3x^2*y^2 = 1,

    3x^2*y^2 = 1 - 1%2F4 = 3%2F4,

    x^2^*y^2 = 1%2F4.


Now   

    1%2Fx%5E6 + 1%2Fy%5E6 = %28x%5E6%2By%5E6%29%2F%28x%5E6%2Ay%5E6%29 = %28%281%2F4%29%29%2F%28%281%2F4%29%5E3%29 = 4%5E3%2F4 = 4%5E2 = 16.


Thus  1%2Fsin%5E6%28beta%29 + 1%2Fcos%5E6%28beta%29 = 16.


ANSWER.   1%2Fsin%5E6%28beta%29 + 1%2Fcos%5E6%28beta%29 = 16.

At this point, the problem is solved completely.