Auxillary angle method of solving trig equations (1 Viewer)

[Faytle]

Hey .
I just started this topic today, and so far I have no clue what I'm doing. I can barely do the questions in the textbook apart from a few here and there. Since there're too many I'll just pick one!

Solve for 0º <= x <= 360º using the auxillary angle method
5cosx - 12sinx = -3

Please and thank you! Any extra tips would also be appreciated

 

[kony]

put it into the form rcos(x+a) = -3

then, expanding, rcosacosx - rsinasinx = -3

equating coefficients, rcosa = 5 and rsina = 12
cosa = 5/r and sina = 12/r

consider a right triangle, with sides 5, 12, r (with r being hypotenuse and angle a adjacent to 5)

from pythagoras', r = 13. and a = inverse tan12/5

that should be enough.

 

[Faytle]

put it into the form rcos(x+a) = -3

then, expanding, rcosacosx - rsinasinx = -3

equating coefficients, rcosa = 5 and rsina = 12
cosa = 5/r and sina = 12/r

consider a right triangle, with sides 5, 12, r (with r being hypotenuse and angle a adjacent to 5)

from pythagoras', r = 13. and a = inverse tan12/5

that should be enough.

Yep I got that far, but its the steps after that that I'm confused about .

 

[SoulSearcher]

Yep I got that far, but its the steps after that that I'm confused about .

Basically, you have 13 cos [x + tan^-1(12/5)] = -3, then solve from there.

Spoiler

13 cos [x + tan^-1(12/5)] = -3
cos [x + tan^-1(12/5)] = -3/13
x + tan^-1(12/5) = cos^-1(-3/13)
x = cos^-1(-3/13) - tan^-1(12/5)

 

[Faytle]

Basically, you have 13 cos [x + tan^-1(12/5)] = -3, then solve from there.

Spoiler

13 cos [x + tan^-1(12/5)] = -3
cos [x + tan^-1(12/5)] = -3/13
x + tan^-1(12/5) = cos^-1(-3/13)
x = cos^-1(-3/13) - tan^-1(12/5)

Ooh righto! In the answers there's also another answer though?

And just wondering if this is an easier way to do it? My teacher/textbook does something slightly different in the last steps, which alters the domain according to the auxillary angle?

And thanks a lot btw!

 

[SoulSearcher]

Ooh righto! In the answers there's also another answer though?

And just wondering if this is an easier way to do it? My teacher/textbook does something slightly different in the last steps, which alters the domain according to the auxillary angle?

And thanks a lot btw!

Hmm, what's the answer they give?

What do you mean? I've probably encountered that method of solution before, but I must've forgotten now.

No problem

 

[Faytle]

The answers are 35°58' and 189°16'.

Teacher explained it today so I get it now. Thanks though!

 

[zelda011]

Another way:

Just like Kony said, use the Sum/Difference expansion to get:

Acos(x+y) = -3
Acosxcosy - Asinxsiny = -3

equating coefficients, Acosy = 5 (label equation 1) and Asiny = 12 (label equation 2)
Now heres where my method differs.

Square equations one and two, then add them together.
A^2 * cos^2y + A^2 * sin^2y = 25 + 144
Take A^2 as a factor of LHS:
A^2 (cos^2y + sin^2y) = 169

Therfore A^2 = 169 (since, from Identities, (cos^2y + sin^2y) = 1)

A = 13

Sub back into eqn 1 or 2, doesnt matter which.

i'll use equation 1.

13siny = 5
siny = 5/13
y = 22.62 degrees.

from there you can work out all the values of Asin(x-y) = -3 (above):
13cos(x-22.62) = -3
cos(x-22.62) = -3/13
x-22.62 = 103.3 degrees (so its really 180 - 103.3 = 76.7 degrees)(i think, its too late for this crap)
x-22.62 =76.7, 360-76.7
therefore add 22.62 to get the angles. You can see its wrong, thats because its 12am and ive probably made 30 calculation errors but you get the method (i hope) which is what counts.