First slide
Multiple and sub- multiple Angles
Question

Let f(θ)=sinθsin3θsin5θ , then f(π/14) is equal to

Moderate
Solution

As π14π23π7,3π14π22π7 

and 5π14=π2π7 , we can write  

 fπ14=cos3π7cos2π7cosπ7 2sinπ7fπ14=sin2π7cos2π7cos3π7 4sinπ7fπ14=sin4π7cos3π7 8sinπ7fπ14=sin7π7+sinπ7=sinπ7 fπ14=18

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