MHB Finding Period, Amplitude, and Phase Shift in Trig Expressions

[karush]

View attachment 1654
first not completely convinced that these answers are correct
I assume the x-axis is in radians.

also if $$2csc(\theta) = 5.633$$ but it does not cross the $$x$$ axis why then does it work in the equation

also, sure there is a better ways to solve this. but using the triangle was the just one way I could think of.

 

[Evgeny.Makarov]

Re: cot(x)+2csc(x)-3=0

I would write the equation using sine and cosine, simplify as much as possible, then express sine as 1 minus cosine squared and solve the quadratic equation to find cosine.

I suspect θ cannot be found precisely, only approximately.

 

[karush]

Re: cot(x)+2csc(x)-3=0

I would write the equation using sine and cosine, simplify as much as possible, then express sine as 1 minus cosine squared and solve the quadratic equation to find cosine.

isn't $\displaystyle\sin^2(\theta) = 1-\cos^2(\theta)$

then $\displaystyle\sin(\theta) = \sqrt{1-\cos^2(\theta)}$

if so then

$
\displaystyle
\frac{\cos{\theta}}{\sqrt{1-\cos^2{\theta}}}
+
\frac{2}{\sqrt{1-\cos^2{\theta}}}
=
\frac{\cos{\theta}+2}{\sqrt{1-\cos^2{\theta}}}
=3
$

this still looks a little daunting

 

[Evgeny.Makarov]

Re: cot(x)+2csc(x)-3=0

isn't $\displaystyle\sin^2(\theta) = 1-\cos^2(\theta)$

Yes, sorry.

$
\displaystyle
\frac{\cos{\theta}}{\sqrt{1-\cos^2{\theta}}}
+
\frac{2}{\sqrt{1-\cos^2{\theta}}}
=
\frac{\cos{\theta}+2}{\sqrt{1-\cos^2{\theta}}}
=3
$

this still looks a little daunting

Don't be daunted. :) Multiply both sides by the denominator, take the square of both sides, denote $\cos\theta$ by $x$ and solve the resulting quadratic equation for $x$.

 

[MarkFL]

Re: cot(x)+2csc(x)-3=0

Another approach would be to multiply through by $\sin(x)$ and arrange as:

$$3\sin(x)-\cos(x)=2$$

Using a linear combination identity, we may write:

$$\sqrt{10}\sin\left(x-\tan^{-1}\left(\frac{1}{3} \right) \right)=2$$

$$\sin\left(x-\tan^{-1}\left(\frac{1}{3} \right) \right)=\frac{2}{\sqrt{10}}=\frac{\sqrt{10}}{5}$$

Now, using the identity $\sin(\pi-\theta)=\sin(\theta)$, can you find the solutions?

 

[karush]

Re: cot(x)+2csc(x)-3=0

$$10cos^2\theta+4cos\theta-5=0$$

so from quadratic formula
$$
\cos\theta
= \frac{-2\pm 3 \sqrt{6}}{10}
=-0.934847 \text{ or }0.534847
$$

so

$$\cos^{-1}(-0.934847)=2.77862 \text{ rad} =159.20^o$$

and

$$\cos^{-1}(0.534847)=1.00647 \text{ rad} =57.67^o$$

check

$$\cot(2.77862)+2\csc(2.77862)-3=0$$
$$\cot(1.00647)+2\csc(1.00647)-3=0$$

 

[MarkFL]

Re: cot(x)+2csc(x)-3=0

Using the method I gave, you find:

$$x=\sin^{-1}\left(\frac{\sqrt{10}}{5} \right)+\tan^{-1}\left(\frac{1}{3} \right)\approx1.00646975739893$$

$$x=\pi-\sin^{-1}\left(\frac{\sqrt{10}}{5} \right)+\tan^{-1}\left(\frac{1}{3} \right)\approx2.77862400498415$$

 

[Evgeny.Makarov]

Re: cot(x)+2csc(x)-3=0

$$10cos^2\theta+4cos\theta-5=0$$

so from quadratic formula
$$
\cos\theta
= \frac{-2\pm\sqrt{6}}{10}
=-0.934847 \text{ or }0.534847
$$

This should be
\[
\cos\theta
= \frac{-2\pm3\sqrt{6}}{10}
\]
but the numerical answers are correct.

 

[karush]

Re: cot(x)+2csc(x)-3=0

thanks for the solutions. learned a lot from this one

one question I have these expressions with multiple trig functions in them... how do you find out the period, amplitude and phase shift on these. or do you just look at a plot..