InterviewSolution
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InterviewSolution
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Evaluate: i) `sin{pi/3-sin^(-1)(-1/2)}` ii) `sin(1/2cos^(-1)4/5)` iii) `sin(cot^(-1)x)` iv) `tan1/2(cos^(-1)sqrt(5)/3)` |
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Answer» i) We know that `sin^(-1)(-theta)=-sin^(-1)theta`. `therefore sin{pi/3-sin^(-1)(-1/2)}=sin{pi/3+sin^(-1)1/2}=sin(pi/3+pi/6)[therefore sin^(-1)1/2=pi/6]` `=sin pi/2=1`. ii) Let `cos^(-1)4/5=theta`, where `theta in [0,pi]`. Then, `costheta=4/5`. Since, `theta in [0,pi] rArr 1/2 theta in [0,pi/2] rArr sin1/2theta gt 0`. `therefore sin(1/2cos^(-1)4/5)=sin1/2theta=sqrt((1-costheta)/(2)) = sqrt((1-(4/5))/(2)=1/sqrt(10)`. ii) Let `cot^(-1)x=theta`. Then, `theta in [0,pi]`. `therefore sin(cot^(-1)x)=sin theta gt 0`. Now, `sin theta=1/("cosec"theta)=1/sqrt(1+cot^(2)theta)=1/sqrt(1+x^(2))`. `therefore sin(cot^(-1)x)=sintheta=1/sqrt(1+x^(2))`. iv) Let `cos^(-1)sqrt(5)/3=theta`. Then, `costheta=sqrt(5)/3`, where `theta in [0,pi]`. `therefore tan1/2(cos^(-1)sqrt(5)/3)=tan1/2theta=sqrt((1-costheta)/(1+costheta))=sqrt((1-sqrt(5)//3)/(1+sqrt(5)//3))=sqrt((3-sqrt(5))/(3+sqrt(5)) xx (3-sqrt(5))/(3-sqrt(5)))=(3-sqrt(5))/(2)`. |
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