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sin(30+x)sin(30-x)=cosx30 are in degrees.

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barr2012 | Student, Undergraduate | eNotes Newbie

Posted February 18, 2012 at 4:16 AM via web

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sin(30+x)sin(30-x)=cosx

30 are in degrees.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted February 18, 2012 at 4:39 AM (Answer #1)

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You need to evaluate `sin(30+x)`  using the formula `sin(alpha+beta) = sin alpha*cos beta + sin beta*cos alpha ` such that:

`sin(30+x) = sin 30*cos x + sin x*cos 30`

You need to substitute `1/2`  for `sin 30 ` and `sqrt3/2`  for `cos 30 ` such that:

`sin(30+x) = (1/2)*cos x + sin x*(sqrt3/2)`

`sin(30-x) = (1/2)*cos x- sin x*(sqrt3/2)`

You need to evaluate the multiplication `sin(30+x)*sin(30-x)`  such that:

`sin(30+x)*sin(30-x) = ((cos x + sqrt3*sin x)(cos x- sqrt3*sin x))/4`

You need to substitute `(cos x + sqrt3*sin x)(cos x - sqrt3*sin x)`  by the difference of squares such that:

`(cos x + sqrt3*sin x)(cos x - sqrt3*sin x) = cos^2 x - 3sin^2 x`

If you need to prove the identity, this last line proves that the expression is not an identity.

If you need to solve for x the equation, then you need to contine to write the equation in terms of cos x only using the basic trigonometric formula `sin^2 x + cos^2 x = 1` .

`cos^2 x - 3sin^2 x = cos x`

`cos^2 x - 3(1- cos^2 x) = cos x`

`cos^2 x-3+ 3cos^2 x - cos x = 0`

`4cos^2 x - cos x - 3 = 0`

You should come up with the substitution cos x = y such that:

`4y^2 - y - 3 = 0`

You need to use quadratic formula such that:

`y_(1,2) = (1+-sqrt(1 + 48))/8`

`y_1 = (1+7)/8 =gt y_1 = 1`

`y_2 = (1-7)/8 = -3/4`

You need to solve for x such that:

`cos x = 1 =gt x = 2n*pi`

`cos x = -3/4 =gt x = +-cos^(-1)(-3/4) + 2n*pi`

Hence, the expression does not represent an identity and the general solutions to equation `sin(30+x)*sin(30-x) = cos x ` are `x = 2n*pi ` and `x = +-cos^(-1)(-3/` 4) + 2n*pi.

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minhthien94 | Student, Undergraduate | (Level 1) eNoter

Posted February 18, 2012 at 6:55 AM (Answer #3)

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Man, sr i was wrong on my last answer. I don't know how to delete it but I post this one to answer your question.

 

we have

     sin(a).sin(b) = [cos(a-b)-cos(a+b)] /2

so

     sin(30+x).sin(30-x) = cos(x)

     [cos(2x)-cos(60)] /2 = cos(x)

     cos(2x) - cos(60) = 2cos(x)

     {2.[cos(x)]^2 - 1} - 2cos(x) - 1/2 = 0

(because cos(2x) = 2.[cos(x)]^2 -1 and cos(60)=1/2 )

     4.[cos(x)]^2 - 4cos(x) -3 = 0 (Times 2 to both side)

     cos(x) = -1/2 or cos(x) = 3/2 (the root of quadratic equation)

     cos(x) = 1-/2 (eliminate cos(x) = 3/2 b/c cos(x) < 1) 

For cos(x) = -1/2, we have    

     x = 2pi/3 + 2npi or x = -2pi/3 + 2npi (where n is any ineteger)     

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