If \(\sqrt {2}\)cos ∝ = cos β + cos³ β

and \(\sqrt {2}\) sin ∝ = sin β + sin³ β

then find sin 2β

\(\frac{2\sqrt {2}}{\sqrt {7}}\)

(\(\sqrt {2}\)cos ∝)² = (cos β)² + (cos³ β)² + 2cos β cos³ β

(\(\sqrt {2}\)sin ∝)²  = (sin β)² + (sin³ β)² + 2sin β sin³ β

⇒ 2cos² ∝ = cos² β + cos⁶ β + 2cos⁴ β

+ 2sin² ∝ = sin² β + sin⁶ β + 2sin⁴ β

---------------------------------------------------------------------

2 × 1   = 1 + 1 - 3sin² β cos² β + 2 (1 - 2sin² β cos² β)

7 sin² β cos² β = 2

multiply both side by 4

= 4 × 7sin² β cos² β = 2 × 4

  ( 2sin β cos β)² = \(\frac{8}{7}\)

             ↓

⇒   sin2 β = \(\sqrt {\frac{8}{7}}\)

⇒   sin2 β = \(\frac{2 \sqrt {2}}{\sqrt {7}}\)