Peyton J.

asked • 04/20/15

General solution of; sin(x-20)=cos(x+30)

How do I find the general solution, do I need to equate them to zero?

Stephanie M.

tutor
This isn't a great answer, so I'm leaving it as a comment so other tutors will still see this question listed as "unanswered." Just wanted to give you my thoughts in case you needed an answer quickly.
 
This problem asks you to figure out where two different angles A = x - 20 and B = x + 30 have the same values for sine and cosine respectively. Remember that, on the unit circle, angles can be represented as points where the x-coordinate is their cosine and the y-coordinate is their sine. For example, 30º is located at (√3/2, 1/2), so cos(30º) = √3/2 and sin(30º) = 1/2.
 
This means that figuring out when sinA = cosB is the same as figuring out when the y-coordinate of one angle is the same as the x-coordinate of another. Stare at the unit circle a bit and you'll see that this happens whenever two angles are reflections across the line y = x (the perfectly-diagonal line through the origin). In other words, A has to be as much more than 0º as B is less than 90º. So, 10º and 80º work. So do 30º and 60º, 45º and 45º, 7º and 83º, etc. In each pair, sinA = cosB. And notice that each pair adds together to be 90º.
 
But your problem has one further restriction. The pair of angles is x-20 and x+30. Since the pair should add together to equal 90º, like those above, you can say that x - 20 + x + 30 = 90. This means that 2x + 10 = 90, or 2x = 80, or x = 40. And for a general solution, you would write x = 40 + 360n.
 
Just one more wrinkle: I really only focused on Quadrant I. There are pairs of angles in Quadrant III that work as well, since the line y = x passes through it. (We can ignore Quadrants II and IV since sine and cosine have different signs there; one is negative and one is positive.) These angle pairs include 180º and 270º, 190º and 260º, 220º and 230º, etc. All these angles add up to be 450º.
 
This means that, like before, x - 20 + x + 30 = 450, so 2x + 10 = 450, so 2x = 440 and x = 220. For a general solution, you would write x = 220 + 360n.
 
This should be all possible solutions, but you can condense those two general equations into one since 220º is exactly 180º (half the unit circle) past 40º. You can write: x = 40º + 180ºn.
 
Hope this helps!
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04/20/15

1 Expert Answer

By:

Raymond B. answered • 12/26/20

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5 (2)

Math, microeconomics or criminal justice

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sin(x-20) = sinxcos20 - sin20cosx

cos(x+30) = cosxcos30 - sinxsin30


so


sinxcos20-sin20cosx = cosxcos30 - sinxsin30


sinxcos20 + sinxsin30 = cosxcos30 + sin20cosx

sinx(cos20+sin30) = cosx(cos30+sin20)


sinx/cosx = tanx = (cos30+sin20)/(cos20+sin30)

tanx = (sqr3/2 + sin20)/(cos20+ 1/2)

x = tan^-1(sqr3/2+sin20/cos20+1/2)



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