Question

\[Find~ the~ value~ of ~~\alpha~~ btween ~{1\over2}\pi\]\[for~which ~~3sinx+2cosx=\sqrt{13}\sin(x+\alpha)\]

Answer 1

\[btween~~ 0 ~and ~{1\over2} \pi\]

Answer 2

\[\tan \alpha=\frac{2}{3}\]

Answer 3

I don't think that's the right answer?

Answer 4

It says 0.588 is the answer...

Answer 5

tan inverse of 2/3 is 0.588..

Answer 6

Yeah it is inverse of tan in radian measure

Answer 7

Oh, yes. I was working in degrees.

Answer 8

How did you get it?

Answer 9

I used a calculator, lol
\[a = \tan^{-1} (\frac{2}{3} )\]

Answer 10

No, lol. I mean tanα=2/3?

Answer 11

Auxilliary Angle..
\[asinx + bcosx = \sqrt{a^2+b^2} \sin(x+\alpha) \]
\[\tan \alpha = \frac{b}{a} \]

Answer 12

Multiply by sqrt(13)
NOW you have 2/sqrt(13) and 3/sqrt(13) beside sinx and cosx
these are sides of a triangle

you get some insight

Answer 13

@Mimi_x3 I don't completely understand you.
@NotSObright My visual thinking is very bad... (in other words~ I can't work with diagrams) lol

Answer 14

What don't you understand ? It's the formula..

Answer 15

I understand the formula... but I don't know how you arrive at tanα=b/a with the formula...

Answer 16

\[\cos(\alpha)=\frac{2}{\sqrt13}\]
\[\sin(\alpha)=\frac{3}{\sqrt13}\]
subtitute this to get the compund angle formula

Answer 17

I can see how tanα=b/a becomes tanα=2/3 and then the answer... but not how asinx+bcosx=√a^22+b^22sin(x+α) ~~~~~~~~~~~~ tanα=b/a?

Answer 18

given a sin (x) + b cos(x) it can be rewritten as

\[r \sin(x + \alpha)\]

\[r = \sqrt{a^2 + b^2}\] pythagorean identity

\[\tan \alpha = b/a\]

in your question \[r = \sqrt{3^2 + 2^2} \]

\[r = \sqrt{13}\]

\[\tan \alpha = 2/3\]

\[\alpha = \arctan (2/3)\]

\[\alpha = 0.588 \] radians

then

\[\sqrt{13} \sin(x + 0.588)\]

Answer 19

Ok, but how did you get tanα=b/a?