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Home / Doubts and Solutions/JEE/Mathematics

Prove that 

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Asked by Anish 11th July 2018, 1:27 PM
Answered by Expert
Answer:
sinθ sin(60º - θ) sin (60º + θ) 
= sinθ (sin60º cosθ - cos60º  sinθ) (sin60º cosθ + cos60º  sinθ) 
begin mathsize 16px style equals sinθ space open parentheses fraction numerator square root of 3 over denominator 2 end fraction cosθ minus 1 half sinθ close parentheses space open parentheses fraction numerator square root of 3 over denominator 2 end fraction cosθ plus 1 half sinθ close parentheses equals sinθ space open parentheses 3 over 4 cos squared straight theta minus 1 fourth sin squared straight theta close parentheses equals sinθ space open parentheses 3 over 4 open parentheses 1 minus sin squared straight theta close parentheses minus 1 fourth sin squared straight theta close parentheses equals sinθ open parentheses 3 over 4 minus 3 over 4 sin squared straight theta minus 1 fourth sin squared straight theta close parentheses equals sinθ open parentheses 3 over 4 minus sin squared straight theta close parentheses equals 1 fourth cross times open parentheses 3 sinθ minus 4 sin cubed straight theta close parentheses equals 1 fourth sin 3 straight theta minus 1 less or equal than sinθ less or equal than 1 rightwards double arrow negative 1 less or equal than sin 3 straight theta less or equal than 1 open vertical bar sinθsin open parentheses 60 degree minus straight theta close parentheses sin open parentheses 60 degree plus straight theta close parentheses close vertical bar less or equal than 1 fourth end style
Answered by Expert 11th July 2018, 4:47 PM
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