document.write( "Question 467094: cos (theta-(3pi/4)) is equivalent to which of the following?\r
\n" ); document.write( "\n" ); document.write( "a. \"expr%28-sqrt%283%29%2F2%29%28cos%28theta%29-sin%28theta%29%29\"
\n" ); document.write( "b. \"expr%28-sqrt%283%29%2F2%29%28cos%28theta%29%2Bsin%28theta%29%29\"
\n" ); document.write( "c. \"expr%28-sqrt%282%29%2F2%29%28cos%28theta%29%2Bsin%28theta%29%29\"
\n" ); document.write( "d. \"cos%28theta%29+-+sqrt%282%29%2F2\"
\n" ); document.write( "e. \"expr%28-sqrt%282%29%2F2%29%28cos%28theta%29-sin%28theta%29%29\"
\n" ); document.write( "f. \"cos%28theta%29+%2B+sqrt%282%29%2F2\"\r
\n" ); document.write( "\n" ); document.write( "Thank you sooooooooo much in advance!!! This question has been driving me bonkers;-)!!!
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Algebra.Com's Answer #320654 by lwsshak3(11628)\"\" \"About 
You can put this solution on YOUR website!
cos (theta-(3pi/4))
\n" ); document.write( "..
\n" ); document.write( "What you have here is a cosine function of the form: y=Acos(Bx-C) with A=amplitude, period=2π/B, and phase shift=C/B
\n" ); document.write( "For given cosine function, y=cos(x-3π/4)
\n" ); document.write( "Amplitude=1
\n" ); document.write( "B=1
\n" ); document.write( "Period=2π/1=2π
\n" ); document.write( "Phase shift=3π/4 to the right
\n" ); document.write( "Ans:
\n" ); document.write( "What you have here is a cosine function whose value varies from +1 to -1 over one period(2π), depending on what theta(x) is.
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