document.write( "Question 551282: I've been having some problems trying to solve this,
\n" ); document.write( "sin(x+pi/4) + sin(x-pi/4) = 1 for the interval [0,2pi]
\n" ); document.write( "I don't really even know where to start, I think it has something to do with co-function identities, but I don't know how to translate that since it's pi/4 and not pi/2...
\n" ); document.write( "Help is greatly appreciated, Thank you for your time\r
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\n" ); document.write( "\n" ); document.write( "sin(x+pi/4) + sin(x-pi/4) = 1 for the interval [0,2pi]
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Algebra.Com's Answer #359603 by lwsshak3(11628)\"\" \"About 
You can put this solution on YOUR website!
I've been having some problems trying to solve this,
\n" ); document.write( "sin(x+pi/4) + sin(x-pi/4) = 1 for the interval [0,2pi]
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\n" ); document.write( "sin(x+pi/4) + sin(x-pi/4) = 1
\n" ); document.write( "Addition formulas for sin:
\n" ); document.write( "sin(s+t)=sin s cos t + cos s sin t
\n" ); document.write( "sin(s-t)=sin s cos t - cos s sin t
\n" ); document.write( "..
\n" ); document.write( "sin(x+π/4)=sin x cos π/4 + cos x sin π/4
\n" ); document.write( "=sin x*√2/2+cos x*√2/2=√2/2 sin x + √2/2 cos x
\n" ); document.write( "..
\n" ); document.write( "sin(x-π/4)=sin x cos π/4 - cos x sin π/4
\n" ); document.write( "=sin x*√2/2-cos x*√2/2=√2/2 sin x - √2/2 cos x
\n" ); document.write( "..
\n" ); document.write( "√2/2 sin x + √2/2 cos x + √2/2 sin x - √2/2 cos x=1
\n" ); document.write( "√2 sin x=1
\n" ); document.write( "sin x=1/√2
\n" ); document.write( "x=π/4\r
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