Question

If sin to the power 4 theta upon a + cos to the power 4 theta upon B is equals to one upon A + B then prove that sin to the power 8 theta upon a cube + cos to the power 8 theta upon b cube is equals to one upon A + B whole cube

Answer 1

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Answer 2

Given that,

$\frac{sin^4\alpha }{a} + \frac{cos^4\alpha }{b} = \frac{1}{a+b}$

We have to prove that

$\frac{sin^8\alpha }{a^3} +\frac{cos^8\alpha }{b^3} = \frac{1}{(a+b)^3}$

• Now, we have the the given equation:

         $\frac{sin^4\alpha }{a} + \frac{cos^4\alpha }{b} = \frac{1}{a+b}$

      Multiplying the above equation by  $ab(a+b)$ , we get

      $b(a+b)sin^4\alpha +a(a+b)cos^4\alpha = ab$

• Now, using the identity

       $cos^2A-1=sin^2A$, we get

      $b(a+b)sin^4\alpha +a(a+b)(1-sin^2\alpha)^2 = ab$

• Expanding $(1-sin^2\alpha )^2$ using the property

      $(a -b)^2=a^2+b^2-2ab$, we get

      $b(a+b)sin^4\alpha +a(a+b)(1+sin^4\alpha -2sin^2\alpha) = ab$

      $(a+b)^2sin^4\alpha -2a(a+b)sin^2\alpha + a^2 = 0$

•  Now this equation is in the form of $a^2+b^2-2ab=(a-b)^2$ where

      $a= (a+b)sin^2\alpha , b=a$, therefore we get

       $((a+b)sin^2\alpha -a)^2=0$

       Now, we get

        $sin^2\alpha = \frac{a}{a+b}$   and $cos^2\alpha =\frac{b}{a+b}$

•  Now,

        $\frac{sin^8\alpha }{a^3} +\frac{cos^8\alpha }{b^3} = \frac{a^4}{(a+b)^4a^3} + \frac{b^4}{(a+b)^4b^3}$

         $\frac{sin^8\alpha }{a^3} +\frac{cos^8\alpha }{b^3} = \frac{a^4b^3+a^3b^4}{(a+b)^4a^3b^3}$

         $\frac{sin^8\alpha }{a^3} +\frac{cos^8\alpha }{b^3} = \frac{a^3b^3(a+b)}{(a+b)^4a^3b^3}$

         $\frac{sin^8\alpha }{a^3} +\frac{cos^8\alpha }{b^3} = \frac{1}{(a+b)^3}$, hence proved.