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Domain and Range of Basic Inverse Trigonometric Functions

If x = sin ...

Question

If $x = sin \frac{2 π}{7} + sin \frac{4 π}{7} + sin \frac{8 π}{7}$ and $y = cos \frac{2 π}{7} + cos \frac{4 π}{7} + cos \frac{8 π}{7}$ then $x^{2} + y^{2} = ?$

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Solution

$x = s i n (\frac{2 π}{7}) + s i n (\frac{4 π}{7}) + s i n (\frac{8 π}{7}) x^{2} = s i n^{2} (\frac{2 π}{7}) + s i n^{2} (\frac{4 π}{7}) + s i n^{2} (\frac{8 π}{7}) + 2 s i n (\frac{2 π}{7}) s i n (\frac{4 π}{7}) + 2 s i n (\frac{4 π}{7}) s i n (\frac{8 π}{7}) + 2 s i n (\frac{8 π}{7}) s i n (\frac{2 π}{7}) y = c o s (\frac{2 π}{7}) + c o s (\frac{4 π}{7}) + c o s (\frac{8 π}{7}) y^{2} = c o s^{2} (\frac{2 π}{7}) + c o s^{2} (\frac{4 π}{7}) + c o s^{2} (\frac{8 π}{7}) + 2 c o s (\frac{2 π}{7}) c o s (\frac{4 π}{7}) + 2 c o s (\frac{4 π}{7}) c o s (\frac{8 π}{7}) + 2 c o s (\frac{8 π}{7}) c o s (\frac{2 π}{7}) S o, x^{2} + y^{2} = (s i n^{2} (\frac{2 π}{7}) + c o s^{2} (\frac{2 π}{7})) + (s i n^{2} (\frac{4 π}{7}) + c o s^{2} (\frac{4 π}{7})) + (s i n^{2} (\frac{8 π}{7}) + c o s^{2} (\frac{8 π}{7})) + 2 (s i n (\frac{2 π}{7}) s i n (\frac{4 π}{7}) + c o s (\frac{2 π}{7}) c o s (\frac{4 π}{7})) + 2 (s i n (\frac{4 π}{7}) s i n (\frac{8 π}{7}) + c o s (\frac{4 π}{7}) c o s (\frac{8 π}{7})) + 2 (s i n (\frac{8 π}{7}) s i n (\frac{2 π}{7}) + c o s (\frac{8 π}{7}) c o s (\frac{2 π}{7})) A^{2} + B^{2} = 1 + 1 + 1 + 2 c o s (\frac{2 π}{7}) + 2 c o s (\frac{4 π}{7}) + 2 c o s (\frac{6 π}{7}) x^{2} + y^{2} = 3 + 2 (c o s (\frac{2 π}{7}) + c o s (\frac{6 π}{7})) + 2 c o s (\frac{4 π}{7}) . A^{2} + B^{2} = 3 + 2 (2 c o s (\frac{4 π}{7}) c o s (\frac{2 π}{7})) + 2 c o s (\frac{4 π}{7}) A^{2} + B^{2} = 3 + 4 c o s (\frac{4 π}{7}) c o s (\frac{2 π}{7}) + 2 c o s (\frac{4 π}{7}) A^{2} + B^{2} = 3 + 2 c o s (\frac{4 π}{7}) (2 c o s (\frac{2 π}{7}) + 1)$

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