SOLUTION: 1. Express 3cos(x°)+5sin(x°) in the form kcos(x°-a°), where k>0 and 0≤a≤360 Can you provide full working with the solutions please. Many Thanks, Andrew.

Algebra ->  Test -> SOLUTION: 1. Express 3cos(x°)+5sin(x°) in the form kcos(x°-a°), where k>0 and 0≤a≤360 Can you provide full working with the solutions please. Many Thanks, Andrew.       Log On


   



Question 388621: 1. Express 3cos(x°)+5sin(x°) in the form kcos(x°-a°), where k>0 and 0≤a≤360
Can you provide full working with the solutions please.
Many Thanks,
Andrew.

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!

3cos%28x%29%2B5sin%28x%29
Draw a right triangle with those coefficients as legs:

horizontal leg = 3 and vertical leg = 5

 

Calculate the hypotenuse using the Pythagorean theorem

c%5E2=a%5E2%2Bb%5E2
c%5E2=3%5E2%2B5%5E2
c%5E2=9%2B25
c%5E2=34
c=sqrt%2834%29

 

sin%28theta%29+=+5%2Fsqrt%2834%29 so 5=sqrt%2834%29sin%28theta%29

cos%28theta%29+=+3%2Fsqrt%2834%29 so 3=sqrt%2834%29cos%28theta%29

Substitute sqrt%2834%29cos%28theta%29 for 3 and
substitute sqrt%2834%29sin%28theta%29 for 5 in


3cos%28x%29%2B5sin%28x%29 =

sqrt%2834%29cos%28theta%29cos%28x%29%2Bsqrt%2834%29sin%28theta%29sin%28x%29 =

factor out sqrt%2834%29 on the left side:

sqrt%2834%29%28cos%28theta%29cos%28x%29%2Bsin%28theta%29sin%28x%29%29 =

Now we remember the formula cos%28alpha-beta%29=cos%28alpha%29cos%28beta%29%2Bsin%28alpha%29sin%28beta%29, and the above can be written:

sqrt%2834%29%28cos%28theta%29cos%28x%29%2Bsin%28theta%29sin%28x%29%29 =

sqrt%2834%29cos%28theta-x%29

Since cos%28alpha%29+=+cos%28-alpha%29, the above becomes:

sqrt%2834%29cos%28-%28theta-x%29%29

sqrt%2834%29cos%28-theta%2Bx%29

sqrt%2834%29cos%28x-theta%29


This is in the form  k%2Acos%28%22x%B0%22-%22a%B0%22%29

where k=sqrt%2834%29 and theta=arctan%285%2F3%29+=+%2259.03624347%B0%22
 
So we have

3cos%28x%29%2B5sin%28x%29 = k%2Acos%28%22x%B0%22-%22a%B0%22%29 = sqrt%2834%29%2Acos%28%22x%B0%22-%2259.03624347%B0%22%29

Edwin