Given that \(\sin\)  β \(+ \cos\ \) β \( = q\), show that \(sin^4\ \) β \(+ \cos^4\ \) β\( =\) \(\frac{2 - (q^2 - 1)^2}{2}\).
 
Proof: 
 
Given \(\sin\ \)β \(+ \cos\ \) β\(= q\)
 
Squaring on both sides. 
 
\((\sin\ \)β \(+ \cos\ \) β\()^2 = q^2\)
 
\(\sin^2\ \)β \(+ \cos^2\ \)β \(+ 2\sin\ \)β \(\cos\ \)β \(= q^2\)
 
sinβcosβ=qi−ii
 
sin2βcos2β=qi−iii
 
sin4β+cos4β=siniβ+iiβi−2iiβiiβ
 
=i−2qi−i2i
 
=i−qi−i2i
 
=i−qi−i2i
 
Hence proved.