# Annoying trig eqn

1. Jan 30, 2008

### ice109

can anyone think of a cute way to solve this for theta or at least make it linear in one of the trig functions?

$$\frac{1}{2} = k\cos(\theta) -\sin(\theta)$$

2. Jan 31, 2008

### sutupidmath

,
well i think this approach would work-i am not sure though-:

$$\frac{1}{2} = k((\cos(\theta) -\sin(\theta)))$$

$$\frac{1}{2}=k\cos(\theta)-(\(1-cos(\theta))^{1/2}$$, now square both sides, and after some transformations we get:

$$\sin(2\theta)=1-\frac{1}{4k^{2}}$$

i think this would do.

Last edited: Jan 31, 2008
3. Jan 31, 2008

### sutupidmath

Ignore it, i thought k was multiplying the whole thing.><

4. Jan 31, 2008

### AFG34

-(sin-inverse((1-2kcos@)/2)) = @
where @ = theta

or

cos-inverse((1+2sin@)/2K) = @
where @ = theta

i could be wrong though

Last edited: Jan 31, 2008
5. Jan 31, 2008

### ice109

anyone else?

6. Jan 31, 2008

### VietDao29

There should be a chapter in your textbook about solving equation of the form:

a sin(x) + b cos(x) = c.

We can just divide both sides by (k2 + 1)1/2, to get:

$$\frac{1}{2\sqrt{k ^ 2 + 1}} = \frac{k}{\sqrt{k ^ 2 + 1}} \cos(\theta) - \frac{1}{\sqrt{k ^ 2 + 1}} \sin(\theta)$$

Now, let angle $$\alpha$$, be an angle such that:

$$\cos \alpha = \frac{k}{\sqrt{k ^ 2 + 1}}$$, and $$\sin \alpha = \frac{1}{\sqrt{k ^ 2 + 1}}$$, such angle does exist, because:

$$\sin ^ 2 \alpha + \cos ^ 2 \alpha = \frac{k ^ 2}{k ^ 2 + 1} + \frac{1}{k ^ 2 + 1} = 1$$

$$\frac{1}{2\sqrt{k ^ 2 + 1}} = \cos(\alpha + \theta)$$, which is easy to solve. Right? :)

7. Jan 31, 2008

### NateTG

You can do the following:
$$\sin=\pm \sqrt{1-\cos^2}$$
so
$$\frac{1}{2}=k \cos \theta \pm \sqrt{1 - (\cos \theta)^2}$$
$$0=k \cos \theta - \frac{1}{2} \pm \sqrt{1-(\cos \theta)^2}$$
multiply both sides by
$$(k \cos \theta - \frac{1}{2}) \mp \sqrt {1-(\cos \theta)^2}$$
Which eliminates the square root, and leads to a quadratic in $\cos \theta$

8. Jan 31, 2008

### ice109

indeed very cute