If cos 50° + sin 50° = K, then what is the value of cos³ 50° - sin³ 50°?

\(\frac{K^2 + 1}{2}\) \(\sqrt {2 - K^2}\)

⇒ cos 50° + sin 50° = K   ......(i)

So, cos 50° - sin 50° = \(\sqrt {2 - K^2}\)

on squaring equation (i)

⇒ 1 + 2cos 50° sin 50° = K²

⇒ cos 50° sin 50° = \(\frac{K^2 - 1}{2}\)

Now,

cos³ 50° - sin³ 50° = (cos 50° - sin 50°) (cos² 50° + sin² 50° + cos 50° sin 50°)

                               = \(\sqrt {2 - K^2}\) × [ 1 + \(\frac{K^2 - 1}{2}\)]

                               = \(\frac{(K^2 + 1)}{2}\) \(\sqrt {2 - K^2}\)