First slide
Multiple and sub- multiple Angles
Question

cos3θcos3θ+sin3θsin3θ is equal to

Moderate
Solution

 

cos3θcos3θ+sin3θsin3θ=4cos3θ3cosθcos3θ+3sinθ4sin3θsin3θ=4×3cos2θsin2θ4sin2θcos2θ=12cos2θsin2θ×1sin2θ

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