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Doctors Medicines MCQs Q&A

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1 Answers

A:
$cos(t)=cos(t/2+t/2)$
$-7//9=cos^2(t/2)-sin^2(t/2)$
$-7/9=cos^2(t/2)-(1-cos^2(t/2))$
$-7/9=2cos^2(t/2)-1$
$-7/9+1=2cos^2(t/2)$
$2/9=2cos^2(t/2)$
$1/9=cos^2(t/2)$
$+-sqrt(1/9)=+-1/3=cos(t/2)$
$cos(t/2)=-1/3$ , it has a -ve value because t/2 lie at 2nd quadrant.

B:
$sin(t)=sin(t/2+t/2)$
$-sqrt32/9=2sin(t/2)cos(t/2)$
$-sqrt32/9=2sin(t/2)(-1/3)$, from answer above $cos(t/2)=-1/3$
$-sqrt32/9=-2/3sin(t/2)$
$-sqrt32/9*-3/2=sin(t/2)$
$-sqrt(16*2)/9*-3/2=sin(t/2)$
$-4sqrt(2)/9*-3/2=sin(t/2)$
$(2sqrt2)/3=sin(t/2)$

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