advanced functions Trigonometry simplifying question.

jsamplonius

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Apr 11, 2015
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I was given this question for review for a test and I could not figure out how to simplify it properly.
sin(pi/2 - x) + sin(pi - x) + sin(3pi/2 - x) + sin (2pi - x)

i had given it a try but it was wrong this is what i had.

=(sin pi/2) - sinx + (sin pi) - (sin 3pi/2) - sinx + (sin 2pi) - sinx

= (sin pi/2) + (sin pi) + (sin 3pi/2) + (sin 2pi) - 4sinx

= (sin 5pi) - (4sinx)

my teacher told me this was wrong and i don't know any other way to do it. please show all the steps and the answer if it is found. thanks.
 
I was given this question for review for a test and I could not figure out how to simplify it properly.
sin(pi/2 - x) + sin(pi - x) + sin(3pi/2 - x) + sin (2pi - x)

i had given it a try but it was wrong this is what i had.

=(sin pi/2) - sinx + (sin pi) - (sin 3pi/2) - sinx + (sin 2pi) - sinx

= (sin pi/2) + (sin pi) + (sin 3pi/2) + (sin 2pi) - 4sinx

= (sin 5pi) - (4sinx)

my teacher told me this was wrong and i don't know any other way to do it. please show all the steps and the answer if it is found. thanks.

The principle you need to use is:

sin(A - B) = sin(A) * cos(B) - sin(B) * cos(A)

so then

sin(π/2 - x) = sin(π/2) * cos(x) - sin(x) * cos(π/2) = 1 * cos(x) - sin(x) * 0 = cos(x)

Using that principle:

sin(π/2 - x) = cos(x) ;

sin(π - x) = sin(x) ;

sin(3π/2 - x) = -cos(x) ;

sin(2π - x) = -sin(x)

Now try again.....
 
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