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:

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

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

$$a\cos \theta + b\sin \theta = \sqrt {a^2 + b^2 } \cos (\theta - \alpha )$$

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:

$${\tan ^{ - 1} \left( { - \frac{b}{a}} \right)}$$

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 $cos(b)= \frac{B}{\sqrt{A^2+ B^2}}.$