document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1088220>Question 1088220</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the max value of sin(x+&#960;/6)+cos(x+&#960;÷6)in the intervals (0,&#960;÷2)is attained at </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #702472 by </B><A HREF=/tutors/aboutme.mpl?userid=htmentor><B>htmentor(1343)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=htmentor><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=htmentor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=702472\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> If all we are looking for is the maximum value, then a shift in x of pi/6 will not change the amplitude, only the phase.<BR>\n" );
document.write( "Therefore, we only need to find the maximum value of the function sin(x) + cos(x)<BR>\n" );
document.write( "The max. value is obtained where df/dx = 0 = cos(x) - sin(x)<BR>\n" );
document.write( "Thus tan(x) = 1 -> x = pi/4<BR>\n" );
document.write( "f(pi/4) = 1/sqrt(2) + 1/sqrt(2) = 2/sqrt(2) = sqrt(2)<BR>\n" );
document.write( "Ans: sqrt(2) </SPAN>\n" );
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