You are given that \(\left( x ^ { 2 } + y ^ { 2 } \right) ^ { 3 } = x ^ { 6 } + 3 x ^ { 4 } y ^ { 2 } + 3 x ^ { 2 } y ^ { 4 } + y ^ { 6 }\).
Hence, or otherwise, prove that \(\sin ^ { 6 } \theta + \cos ^ { 6 } \theta = 1 - \frac { 3 } { 4 } \sin ^ { 2 } 2 \theta\) for all values of \(\theta\).
Use the result from part (a) to determine the minimum value of \(\sin ^ { 6 } \theta + \cos ^ { 6 } \theta\).
The questions in this section refer to the article on the Insert. You should read the article before attempting the questions.