If \(\sqrt {2}\)cos ∝ = cos β + cos3 β
and \(\sqrt {2}\) sin ∝ = sin β + sin3 β
then find sin 2β
Answer & explanation
Correct answer: option 2
(\(\sqrt {2}\)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
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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 = \(\frac{8}{7}\)
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⇒ sin2 β = \(\sqrt {\frac{8}{7}}\)
⇒ sin2 β = \(\frac{2 \sqrt {2}}{\sqrt {7}}\)