How can the second form of trigonometric equations be derived?

[danago]

Hey. The latest thing we've been studying in trig classes is equations of the form:

<br /> a\cos \theta + b\sin \theta = c<br />

The book explains that to solve an equation like this, i should write it in the form:

<br /> a\cos \theta + b\sin \theta = \sqrt {a^2 + b^2 } \cos (\theta - \alpha )<br />

Now. The teacher has tried all she can to explain how the second form can be derived, but nobody seems to understand. Could someone please go through it?

Thanks,
Dan.

 

[benorin]

It seems easier to solve such equations this way:

a\cos \theta + b\sin \theta = c

recall that the Pythagorean identity gives \sin\theta =\pm\sqrt{1-\cos^{2}\theta} so substitute this into get

a\cos \theta \pm b\sqrt{1-\cos^{2}\theta} = c

rearrange to get

\pm b\sqrt{1-\cos^{2}\theta} = c-a\cos \theta

square both sides

b^2(1-\cos^{2}\theta) = c^2-2ac\cos \theta+a^2\cos^{2}\theta

cleaning it up gives

(a^2+b^2)\cos^{2}\theta -2ac\cos\theta+c^2-b^2=0

which is a quadratic equation in \cos\theta so use the quadratic formula to solve for

\cos\theta = \frac{2ac\pm\sqrt{4a^2c^2-4(a^2+b^2)(c^2-b^2)}}{2(a^2+b^2)}= \frac{ac\pm b\sqrt{a^2+b^2-c^2}}{a^2+b^2}

which is messy, but it works.

 

[Integral]

Dango,
I have 2 questions:
What happened to the c in the first equation?

Where did the alpha in the second come from?

 

[Parlyne]

The idea behind this rewriting is that by you can always make \sqrt {a^2 + b^2 } \cos (\theta - \alpha ) equal to a\cos \theta + b\sin \theta by choosing \alpha correctly. We can see this by considering the angle addition formula for cosines:

\cos (\theta + \phi) = \cos \theta \cos \phi - \sin \theta \sin \phi

If we replace \phi with -\alpha, we can expand the right hand side of the equation and find the value of \alpha that makes the equation true (i.e. that makes the coefficients of \sin \theta and \cos \theta correct).

 

[pavadrin]

\alpha is an obtuse angle which is equal to:

<br /> {\tan ^{ - 1} \left( { - \frac{b}{a}} \right)}<br />

as for the rule maybe this link will help you http://mathworld.wolfram.com/HarmonicAdditionTheorem.html
(credit for this link is to be given to J77)

this rule is reffered to as the harmonic addition formula

hope this helps

 

[HallsofIvy]

sin(a- b)= sin(a)cos(b)- cos(a)sin(b).

If you are given Asin(x)- Bcos(x), you can take a= x and then try to find b such that sin(b)= A, cos(b)= B. Of course that will only work if A and B are between -1 and 1 and A²+ B²= 1. Okay, for general A and B, factor out something so that is true: write A sin(x)- B cos(x) as
\sqrt{A^2+ B^2}(\frac{A}{\sqrt{A^2+ B^2}}cos(x)- \frac{B}{\sqrt{A^2+ B^2}}sin(x)
Then if sin(b)= \frac{A}{\sqrt{A^2+ B^2}}, it follows that cos(b)= \frac{B}{\sqrt{A^2+ B^2}}.