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Given cos (x+y)-1=0, verify if sin(2x+y)-sinx=0?

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chezmena | Student, College Freshman | eNoter

Posted January 16, 2011 at 4:50 PM via web

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Given cos (x+y)-1=0, verify if sin(2x+y)-sinx=0?

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giorgiana1976 | College Teacher | Valedictorian

Posted January 16, 2011 at 4:57 PM (Answer #1)

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We'll start by expanding the cosine of the sum:

cos (x+y) = cos x*cos y - sin x*sin y

From enunciation, we know that:

cos x*cos y - sin x*sin y - 1 = 0

We'll add sin x*sin y + 1 both sides:

cos x*cos y = sin x*sin y + 1 (1)

Now, we'll expand the function sin (2x+y):

sin (2x+y) = sin 2x*cos y + sin y*cos 2x

We'll re-write the factor sin 2x:

sin 2x = sin(x+x) = 2sin x*cos x

We'll re-write the factor cos 2x:

cos 2x  = cos (x+x) = 1 - 2(sin x)^2

We'll re-write the sum:

sin (2x+y) = 2sin x*cos x*cos y + sin y*[1 - 2(sin x)^2]

We'll substitute the product cos x*cos y by (1):

sin (2x+y) = 2sin x*(1 + sin x*sin y) + sin y*[1 - 2(sin x)^2]

We'll remove the brackets:

sin (2x+y) = 2sin x + 2(sin x)^2*sin y + sin y -  2(sin x)^2*sin y

sin (2x+y) - 2sin x - sin y = 0

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