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If `sin2 theta=3/4, then sin^(3)theta+cos^(3)theta=`A. `(sqrt5)/(8)`B. `(sqrt7)/(8)`C. `(sqrt11)/(8)`D. none of these |
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Answer» Correct Answer - D We have, `sin^(3)theta+cos^(3)theta=(sintheta+cos theta)(1-(sin2 theta)/(2))` `=sqrt(1+sin2 theta)(1-(sin2 theta)/(2))` `=sqrt(1+(3)/(4))(1-(3)/(8))=(sqrt7)/(2)xx5/8=(5sqrt7)/(16)` |
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