Another Trig EquationDate: 3/30/96 at 6:19:53 From: Anonymous Subject: trigonometry 4sinx + 3cosx = 4 I can't solve this problem at all. Date: 4/1/96 at 10:4:6 From: Doctor Sebastien Subject: Re: trigonometry Hi, First, you must recall the identity sin x cos y + cos x sin y = sin (x + y) The method that I would use to solve this problem uses the above identity. We need to change the form of the lefthand side of the equation to the above form and then solve it. 4 sin x + 3 cos x = 4 Would it not be great if cos y were equal to 4 over an integer, call it R, and sin y = 3 over the same integer, R? Then R {(4/R) sin x + (3/R) cos x} = 4 A clever way to choose R is to make it the hypotenuse of a triangle that has an angle y Thus, R = SQRT (4^2 + 3^2), where SQRT means 'square root' and ^ means 'to the power of'. R = 5. sin y = 3/5 cos y = 4/5 Therefore, y = 36.9 degrees 5 {(4/5) sin x + (3/5) cos x} = 4 (4/5) sin x + (3/5) cos x = 4/5 cos y sin x + sin y cos x = 4/5 sin x cos y + cos x sin y = 4/5 From the identity stated above, sin (x + y) = 4/5 x + y = 53.1 degrees x + 36.9 = 53.1 x = 16.3 degrees I know that these types of problems sometimes require more than one value for x. It would be easy to find other values of x by looking at the range where the sine of an angle is positive and assigning x + y to be equal to that angle. This problem has been solved using one of the many identities that exist. Here are some of those identities: sin (x - y) = sin x cos y - cos x sin y cos (x + y) = cos x cos y - sin x sin y cos (x - y) = cos x cos y + sin x sin y -Doctor Sebastien, The Math Forum |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/