SOLUTION: 21cos(x) + 20 sin(x) can be written in the form Asin(x+ b) where A>0 and -pi < b < pi Thanks already!

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Question 348596: 21cos(x) + 20 sin(x) can be written in the form Asin(x+ b) where A>0 and
-pi < b < pi
Thanks already!

Answer by Edwin McCravy(20081) About Me  (Show Source):
You can put this solution on YOUR website!
21cos%28x%29%2B20sin%28x%29 can be written in the form Asin%28x%2Bb%29 where A>0 and -pi%3Cb%3Cpi
Cheers in advance
Assume:

A%2Asin%28x%2Bb%29+=+21cos%28x%29%2B20sin%28x%29

A%28sin%28x%29cos%28b%29+%2B+cos%28x%29sin%28b%29%29+=+21cos%28x%29%2B20sin%28x%29

Let x+=+0

A%28sin%280%29cos%28b%29+%2B+cos%280%29sin%28b%29%29+=+21cos%280%29%2B20sin%280%29

A%280%2Acos%28b%29%2B1%2Asin%28b%29%29+=+21%281%29%2B20%280%29

A%280%2Bsin%28b%29%29=21

Asin%28b%29+=+21

--------------------------------------
Now go back to:

A%28sin%28x%29cos%28b%29+%2B+cos%28x%29sin%28b%29%29+=+21cos%28x%29%2B20sin%28x%29

And now let x+=+pi%2F2



A%281%2Acos%28b%29%2B0%2Asin%28b%29%29+=+21%280%29%2B20%281%29

A%28cos%28b%29%2B0%29=0%2B20

Acos%28b%29+=+20

-------------------------

So we have this system of equations:

system%28Asin%28b%29+=+21%2C+Acos%28b%29+=+20%29

Dividing equals by equals:

%28A%28sin%28b%29%29%29%2F%28A%28cos%28b%29%29%29=21%2F20

%28cross%28A%29%28sin%28b%29%29%29%2F%28cross%28A%29%28cos%28b%29%29%29=21%2F20

sin%28b%29%2Fcos%28b%29=21%2F20

tan%28b%29=21%2F20

b=%22tan%22%5E%28-1%29%2821%2F20%29 


Draw a right triangle with horizontal leg 20 and vertical leg 21,
so the angle at the bottom left will have tangent 21%2F20, and
therefore be equal to b. 

 and by the Pythagorean theorem 

Since

Asin%28b%29+=+21

A%2821%2F29%29=21

21A+=+21%2A29

A+=+%2821%2A29%29%2F21

A+=+%28cross%2821%29%2A29%29%2Fcross%2821%29

A+=+29

Therefore  A=29 and b=%22tan%22%5E%28-1%29%2821%2F20%29

and:

21cos%28x%29%2B20sin%28x%29=A%2Asin%28x%2Bb%29
21cos%28x%29%2B20sin%28x%29%22%22=%22%2229%2A%22sin%28x%22%22%22%2B%22%22%22tan%22%5E%28-1%29%2921%2F20%22%29%22

Edwin