Question

How would I solve the following trig question

4cosx+3sinx into the form c sin(x-a)

Answer 1

Let:
\[r \cdot \sin(a) = 4\]

\[r \cdot \cos(a) = 3\]

\[\implies r \cdot \sin(a) .\cdot \cos(x) + r \cdot \cos(a) \cdot \sin(x) \implies r(\sin(a) \cos(x) + \cos(a) \sin(x))\]

\[\implies \color{green}{r \cdot \sin(x + a)}\]

Answer 2

We require negative in middle..

Answer 3

I think c is 5

Answer 4

or r

Answer 5

r is 5 here..

\[r^2 = \sqrt{4^2 + 3^2} \implies r = 5\]

Answer 6

The issue I am having is finding A

Answer 7

Divide both..

Answer 8

I know in my book is say a is 5.3559 but I am not sure how.

Answer 9

\[\frac{r \cdot \sin(a)}{r \cdot \cos(a)} = = \tan(a) = \frac{4}{3} \implies a = \tan^{-1}(\frac{4}{3})\]

Answer 10

I get .9273 something must be wrong.

Answer 11

I got 53.13..

Answer 12

Yeah I think because my calculator is in radians.

Answer 13

in degrees I got 53.13 too.

Answer 14

It is in radians and not degrees..

Answer 15

Yes the answer is asked for in radians.

Answer 16

So this is the answer in my opinion if I have done right..

And this is in radians..

Answer 17

Maybe my book has a typo or something because your answer makes sense.

Answer 18

Yes there may be a misprint..

Answer 19

well thanks anyway.

Answer 20

Welcome..

Answer 21

I would start by noting
sin(x-a)= sin(x) cos(a) - cos(x) sin(a)
and
c sin(x-a) is
c cos(a) sin(x) + -c sin(a) cos(x)

match the coefficients to
3 sin(x) + 4 cos(x)

we get c cos(a)= 3 and -c sin(a)= 4
or tan(a)= -4/3
the inverse tan of -4/3 is -53.13 degrees or rather 306.87 degrees or 5.356 rads

also, we get c from
(c cos(a))^2 + ( -c sin(a))^2= 9+16= 25
c^2 ( cos^2(a) + sin^2(a))= 25
c= 5

we get
3 sin(x) + 4 cos(x)= 5 sin(x-5.356)

Answer 22

I am wrong.. @phi

Answer 23

no, you said
"We require negative in middle.."

you just got side tracked...