document.write( "Question 979116: If sin(Θ) = 3/5, and Θ is in quadrant II. determine in exact form:
\n" ); document.write( "sin(Θ + π/6) \n" ); document.write( "

Algebra.Com's Answer #600504 by KMST(5328) $\"\"$ $\"About$

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$\"sin%28theta%29=3%2F5\"$
\n" ); document.write( " $\"theta\"$ is in quadrant II ---> $\"cos%28theta%29%3C0\"$
\n" ); document.write( "We all remember the trigonometric identity
\n" ); document.write( " $\"cos%5E2%28theta%29%2Bsin%5E2%28theta%29=1\"$ , so we know that
\n" ); document.write( " $\"cos%5E2%28theta%29%2B%283%2F5%29%5E2=1\"$ ---> $\"cos%5E2%28theta%29%2B9%2F25=1\"$ ---> $\"cos%5E2%28theta%29=1-9%2F25\"$ ---> $\"cos%5E2%28theta%29=16%2F25\"$
\n" ); document.write( "Then, $\"system%28cos%28theta%29%3C0%2Ccos%5E2%28theta%29=16%2F25%29\"$ ---> $\"cos%28theta%29=-sqrt%2816%2F25%29\"$ ---> $\"cos%28theta%29=-4%2F5\"$
\n" ); document.write( "Many of us remember that $\"sin%28pi%2F6%29=1%2F2\"$ and $\"cos%28pi%2F6%29=sqrt%283%29%2F2\"$ ,
\n" ); document.write( "and a few of us may remember the trigonometric identity
\n" ); document.write( " $\"sin%28A%2BB%29=sin%28A%29cos%28B%29%2Bsin%28B%29cos%28A%29\"$ .
\n" ); document.write( "Putting together all that we now know,
\n" ); document.write( " $\"sin%28theta%2Bpi%2F6%29=sin%28theta%29cos%28pi%2F6%29%2Bsin%28pi%2F6%29cos%28theta%29\"$
\n" ); document.write( " $\"sin%28theta%2Bpi%2F6%29=%283%2F5%29%28sqrt%283%29%2F2%29%2B%281%2F2%29%28-4%2F5%29\"$
\n" ); document.write( " $\"sin%28theta%2Bpi%2F6%29=3sqrt%283%29%2F10-4%2F10%29\"$
\n" ); document.write( " $\"highlight%28sin%28theta%2Bpi%2F6%29=%283sqrt%283%29-4%29%2F10%29\"$ \n" ); document.write( "

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