Solve Annoying Trig Equation for Theta: Linear Solution Using Trig Functions

[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)

 

[sutupidmath]

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)

,
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.

 

[sutupidmath]

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

 

[AFG34]

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

or

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

i could be wrong though

 

[ice109]

anyone else?

 

[VietDao29]

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)

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

Now, your equation will become:

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

 

[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

 

[ice109]

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

Now, your equation will become:

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

indeed very cute