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Evaluate: sin78°cos18°-cos78°sin 18°

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Evaluate: sin78°cos18°-cos78°sin 18°

    • Posted by Sejal Koundal 7 years, 6 months ago

    CBSE > Class 11 > Mathematics
    • 2 answers

    Nancy Rajput 7 years, 6 months ago

    given equation is in the form sin A ×cos B-cosA×sin B=sin(A-B) given A=78 B=18 =Sin(78-18) =sin(60) =root3/2

    1Thank You

    Shreya Singh 7 years, 6 months ago

    We know that, sinA cosB - cosA sinB = sin(A-B).Here,A=78° and B=18°; therefore sin78° cos18°- cos78° sin18° = sin(78°-18°) = sin 60° =√3/2

    3Thank You

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