First slide
Multiple and sub- multiple Angles
Question

16cosθcosπ8cosθcos3π8cosθcos5π8cosθcos7π8=λcos4θ then the value of λ is

Moderate
Solution

16cosθcosπ8cosθcos3π8cosθcos5π8cosθcos7π8

=16cosθcosπ8cosθcos7π8×cosθcos3π8cosθcos5π8

=16cosθcosπ8cosθ+cosπ8×cosθcos3π8cosθ+cos3π8

=16cos2θcos2π8cos2θcos23π8=16cos2θcos2π8cos2θsin2π8=16cos4θcos2θ+sin2π8cos2π8=16cos4θcos2θ+18=16cos2θsin2θ+18=16sin22θ4+18=1612sin22θ8=16cos4θ8

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