Trigonometry Question: (2 Viewers)

Divinity_ · Jul 12, 2009

Solve:

3cos theta - 4sin theta = 5 for -pi <= theta <= pi

tommykins · Jul 12, 2009

use the auxillary angle method, ie. change to Rcos(a+b)

gurmies · Jul 12, 2009

Or t-method, where t = tan(x/2). If this means nothing to you, I can post solution.

Official · Jul 13, 2009

$3cos\Theta -4sin\Theta = 5$

$3cos\Theta -4sin\Theta = \sqrt{\alpha ^2 + \beta ^2}cos(\Theta +\alpha )$

$\sqrt{\alpha ^2 + \beta ^2} = \sqrt{3^2 + 4^2}$

$\sqrt{25}cos(\Theta +\beta ) = 5cos\Theta cos\alpha -5sin\Theta sin\alpha$

Equating coefficients

$cos\alpha = \frac{3}{5}$

$\alpha =53^\circ 8'$

Substituting values back in

$3cos\Theta -4sin\Theta =5cos(\Theta +53^\circ 8' )$

$5cos(\Theta +53^\circ 8' )=5$

$cos(\Theta +53^\circ 8' )=1$

Then you solve this simplified equation for answer.

inhinyero · Jul 13, 2009

transform to polar coordinates.. it would be easier.

Drongoski · Jul 13, 2009

or you can do it this way:

$\80dpi 3cos\theta \ -\ 4sin\theta \ \ =\ \ 5\\ \\ \therefore\ \ 5\ +\ 4sin\theta \ \ =\ \ 3cos\theta\\ \\ squaring\ both\ sides:\\ \\ 16\ sin^{2}\theta\ +40sin\theta +25\ \ =\ \ 9\ cos^{2} \theta \ \ =\ \ 9-9\ sin^{2} \theta\\ \\ \therefore \ \ 25\ sin^{2}\theta +40sin\theta +16\ \ =\ \ 0\\ \\ i.e.\ \ (5\ sin\theta +4)^{2} \ =\ 0\\ \\ \therefore \ \ 5\ sin\theta +4\ =\ 0\ \ \Rightarrow \ \ sin\theta \ =\ -0.8\\ \\ \therefore \ \ \theta \ =\ -53^{\circ} \ 8{}'\ \ \ or\ \ \ -126^{\circ}\ 52{}'\ (exclude\ this)$

Is this answer right ??

edit

2nd answer -126. ... deg should be excluded. Thks Timothy for alerting me.

annabackwards · Jul 13, 2009

Drongoski said:
or you can do it this way:

$\80dpi 3cos\theta \ -\ 4sin\theta \ \ =\ \ 5\\ \\ \therefore\ \ 5\ +\ 4sin\theta \ \ =\ \ 3cos\theta\\ \\ squaring\ both\ sides:\\ \\ 16\ sin^{2}\theta\ +40sin\theta +25\ \ =\ \ 9\ cos^{2} \theta \ \ =\ \ 9-9\ sin^{2} \theta\\ \\ \therefore \ \ 25\ sin^{2}\theta +40sin\theta +16\ \ =\ \ 0\\ \\ i.e.\ \ (5\ sin\theta +4)^{2} \ =\ 0\\ \\ \therefore \ \ 5\ sin\theta +4\ =\ 0\ \ \Rightarrow \ \ sin\theta \ =\ -0.8\\ \\ \therefore \ \ \theta \ =\ -53^{\circ} \ 8{}'\ \ \ or\ \ \ -126^{\circ}\ 52{}'$

I was just about to suggest this method

Drongoski · Jul 13, 2009

annabackwards said:
I was just about to suggest this method

It is often easier and more intuitive but is not widely adopted.

philly2000 · Jul 13, 2009

wow just learnt a new method, thanks drongoski !

Official · Jul 13, 2009

Drongoski said:
or you can do it this way:

$\80dpi 3cos\theta \ -\ 4sin\theta \ \ =\ \ 5\\ \\ \therefore\ \ 5\ +\ 4sin\theta \ \ =\ \ 3cos\theta\\ \\ squaring\ both\ sides:\\ \\ 16\ sin^{2}\theta\ +40sin\theta +25\ \ =\ \ 9\ cos^{2} \theta \ \ =\ \ 9-9\ sin^{2} \theta\\ \\ \therefore \ \ 25\ sin^{2}\theta +40sin\theta +16\ \ =\ \ 0\\ \\ i.e.\ \ (5\ sin\theta +4)^{2} \ =\ 0\\ \\ \therefore \ \ 5\ sin\theta +4\ =\ 0\ \ \Rightarrow \ \ sin\theta \ =\ -0.8\\ \\ \therefore \ \ \theta \ =\ -53^{\circ} \ 8{}'\ \ \ or\ \ \ -126^{\circ}\ 52{}'$

Is this answer right ??

lol this is such a good method! You dont have to go through all that random crap lollzz

Aquawhite · Jul 13, 2009

Whatever you can do to make it simplier... by squaring it makes things so much more simple. I really like that squaring method (very simple and just obvious).

untouchablecuz · Jul 14, 2009

remember squaring can add false solutions

annabackwards · Jul 14, 2009

untouchablecuz said:
remember squaring can add false solutions

How so? It has always worked for me...

khorne · Jul 14, 2009

Well

x = 5
x^2 = 25
sqrt both sides
therefore
x = +/- 5

You get two solutions when you originally had one.

untouchablecuz · Jul 14, 2009

consider

sin x + cos x = 1, 0<=x<=2pi

squaring both sides

sin^2x +2sinxcosx + cos^2x = 1

1 + sin2x = 1

sin2x = 0, 0<=2x<=2pi

solutions: x = 0, pi/2, pi, 3pi/2, 2pi

sub in x = 0, pi/2, 2pi into sin x + cos x = 1 and the equation is satisified

sub in x = pi, 3pi/2 into sin x + cos x and we get -1 =/= 1

hence pi and 3pi/2 are not solutions and were included when squaring

gurmies · Jul 14, 2009

untouchablecuz said:
consider

sin x + cos x = 1, 0<=x<=2pi

squaring both sides

sin^2x +2sinxcosx + cos^2x = 1

1 + sin2x = 1

sin2x = 0, 0<=2x<=2pi

solutions: x = 0, pi/2, pi, 3pi/2, 2pi

sub in x = 0, pi/2, 2pi into sin x + cos x = 1 and the equation is satisified

sub in x = pi, 3pi/2 into sin x + cos x and we get -1 =/= 1

hence pi and 3pi/2 are not solutions and were included when squaring

Exactly. Test when squaring.

annabackwards · Jul 14, 2009

gurmies said:
Exactly. Test when squaring.

Ah, noted ^^

Drongoski · Jul 14, 2009

annabackwards said:
Ah, noted ^^

Caution using the quadratic eqn method

I was going to post a cautionary note about the method I used:

1) don't use it if reqd to do using auxillary angle or t-formula method

2) when you square: LHS = RHS you get new eqn: LHS² = RHS²
The solutions to this new eqn include those for LHS = -RHS.
So you have to test for and to exclude these.

I think it is still a lot shorter and easier in general.

Timothy.Siu · Jul 14, 2009

Divinity_ said:
Solve:

3cos theta - 4sin theta = 5 for -pi <= theta <= pi

5cos(@+tan^-1(4/3))=5

cos(@+tan^-1(4/3))=1
@+tan^-1(4/3)=0
@=-tan^-1(4/3)

drongonski, ur other solution is wrong (i think)
because 3cos -126 would be negative (not in the positive quadrants), and 4sin (anything) can't be greater than 4

Dumbledore · Jul 14, 2009

just wondering when using the acos(x)+bsin(x)=Rsin(x+@)
can u just say R=sqrt(a^2+b^2) and @=arctan(b/a)
or do u have to expand Rsin(x+@) and equate co efficients?

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