Question

Sin^8g/a^3+cos^8y/b^3=1/(a+b)^3

Answer 1

Question:

As far as I can understand, the question was about simplifying the given equation for further solving.

Given equation is:

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

Answer:

The first step we have to do is applying the power-reduction formula:

$$\sin^ {2} \alpha = \frac {1 - \cos 2 \alpha}{2},$$ $$\cos^ {2} \alpha = \frac {1 + \cos 2 \alpha}{2}.$$

In our case:

$$\sin^ {8} y = (\sin^ {2} y) ^ {4} = \left(\frac {1 - \cos 2 y}{2}\right) ^ {4},$$ $$\cos^ {8} y = (\cos^ {2} y) ^ {4} = \left(\frac {1 + \cos 2 \alpha}{2}\right) ^ {4}.$$

On the next step, we change the variable. Let:

$$z = \frac {1 - \cos 2 y}{2}.$$

Now, the given equation turns to:

$$\frac {z ^ {4}}{a ^ {3}} + \frac {(1 - z) ^ {4}}{b ^ {3}} = \frac {1}{(a + b) ^ {3}},$$

which can be transformed to a quartic equation:

$$z ^ {4} - z ^ {3} \frac {4 a ^ {3}}{a ^ {3} + b ^ {3}} + z ^ {2} \frac {6 a ^ {3}}{a ^ {3} + b ^ {3}} - z \frac {4 a ^ {3}}{a ^ {3} + b ^ {3}} + \frac {a ^ {3}}{a ^ {3} + b ^ {3}} - \frac {a ^ {3} b ^ {3}}{(a + b) ^ {3}} = 0,$$

or:

$$z ^ {4} - z ^ {3} \frac {4}{1 + c ^ {3}} + z ^ {2} \frac {6}{1 + c ^ {3}} - z \frac {4}{1 + c ^ {3}} + \frac {1}{1 + c ^ {3}} - \frac {b ^ {3}}{(1 + c) ^ {3}} = 0,$$

where $c = b / a$.

Equation (2) can be solved as a quartic equation.

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