Solving trigonometric equations.

Hello

The question was:

Solve the following on the interval $\displaystyle 0\leq x \leq 2\pi$. Give your answers to 3 sf.

$\displaystyle 2\cos\theta - \sin\theta = 1$

I was able to solve this using the harmonic form, expressing the $\displaystyle LHS$ in terms of $\displaystyle R\cos(\theta + \alpha)$. I would like to know if it is possible, better, to solve this using identities, needless to say I can't see how and would appreciate a hint as how to start.

Thank you.

Re: Solving trigonometric equations.

Quote:

Originally Posted by

**Furyan** Hello

The question was:

Solve the following on the interval $\displaystyle 0\leq x \leq 2\pi$. Give your answers to 3 sf.

$\displaystyle 2\cos\theta - \sin\theta = 1$

I was able to solve this using the harmonic form, expressing the $\displaystyle LHS$ in terms of $\displaystyle R\cos(\theta + \alpha)$. I would like to know if it is possible, better, to solve this using identities, needless to say I can't see how and would appreciate a hint as how to start.

Thank you.

$\displaystyle 2\cos{t} = \sin{t} + 1$

$\displaystyle 4\cos^2{t} = \sin^2{t} + 2\sin{t} + 1$

$\displaystyle 4(1 - \sin^2{t}) = \sin^2{t} + 2\sin{t} + 1$

$\displaystyle 0 = 5\sin^2{t} + 2\sin{t} - 3$

$\displaystyle 0 = (5\sin{t} - 3)(\sin{t} + 1)$

can you finish? don't forget to check for extraneous roots.

Re: Solving trigonometric equations.

Hi Furyan! :)

You would need the right trigonometric identity:

List of trigonometric identities - Linear combinations - Wikipedia

Edit: Oh never mind, I'm guessing that this is what you meant with the harmonic form.

Re: Solving trigonometric equations.

Hello Skeeter

Quote:

Originally Posted by

**skeeter** $\displaystyle 2\cos{t} = \sin{t} + 1$

$\displaystyle 4\cos^2{t} = \sin^2{t} + 2\sin{t} + 1$

$\displaystyle 4(1 - \sin^2{t}) = \sin^2{t} + 2\sin{t} + 1$

$\displaystyle 0 = 5\sin^2{t} + 2\sin{t} - 3$

$\displaystyle 0 = (5\sin{t} - 3)(\sin{t} + 1)$

can you finish? don't forget to check for extraneous roots.

Thank you very much for that. I'll have to look very closely at how you got that quadratic in $\displaystyle \sin t$. Edit: It was the first step I couldn't see, but it's squaring both sides. Now I've seen that it's very much easier than the method I used, which I can really only do with the book in front of me.

As for finishing $\displaystyle \sin t = \dfrac{3}{5}$ gives me $\displaystyle t = 0.644$ and $\displaystyle t = 2.50$ and $\displaystyle \sin t = -1$ gives me $\displaystyle t = 4.71 = (\dfrac{3}{2}\pi)$. The first and the last of these solutions I got using the harmonic form, which, since you mentioned them, leads me to believe that $\displaystyle t = 2.50$ is an extraneous root, but looking at the graph of $\displaystyle y = \sin x$ I don't see why. Could you please explain that.

Thank you very much.

Edit: I just checked the roots using a calculator and found that $\displaystyle t = 2.50$ is not a solution. Is that how I check for extraneous roots, by seeing if they work? :)

Re: Solving trigonometric equations.

Hello ILikeSerena

Quote:

Originally Posted by

**ILikeSerena**

Thank you. Yes I think that's what my book calls the 'harmonic form'. I was hoping there was an easier way, but maybe not. Thanks for your efforts all the same. :)

Re: Solving trigonometric equations.

To answer your question about extraneous roots, substitution into the original equation will in general show if the solution is extraneous or not.

I would have solved the equation much the same way you did:

$\displaystyle 2\cos\theta-\sin\theta=1$

$\displaystyle \sqrt{5}\sin\left(\theta+\pi-\tan^{-1}(2)\right)=1$

$\displaystyle \sin\left(\theta+\pi-\tan^{-1}(2)\right)=\frac{1}{\sqrt{5}}$

First solution:

$\displaystyle \theta+\pi-\tan^{-1}(2)=\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)$

$\displaystyle \theta=\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)+\tan^{-1}(2)-\pi$

Observing that in a $\displaystyle 1,2,\sqrt{5}$ right triangle, the two inverse trig. functions represent complementary angles, we have:

$\displaystyle \theta=\frac{\pi}{2}-\pi=-\frac{\pi}{2}$

Adding $\displaystyle 2\pi$ to get the angle in the desired range, we get:

$\displaystyle \theta=-\frac{\pi}{2}+2\pi=\frac{3\pi}{2}$

Second solution:

$\displaystyle \theta+\pi-\tan^{-1}(2)=\pi-\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)$

$\displaystyle \theta=\tan^{-1}(2)-\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)$

We could plug this into a calculator, but there is a slightly simpler representation.

Construct a right triangle with horizontal base 1, altitude 2 and hypotenuse $\displaystyle \sqrt{5}$ with the right angle at the lower right, then from the lower left vertex draw a line segment to the vertical leg at the point 1/2 unit above the right angle, and we have the angle subtended by this segment and the hypotenuse as $\displaystyle \theta$. Using the law of cosines, we find:

$\displaystyle \left(\frac{3}{2}\right)^2=\left(1^2+\left\(\frac{ 1}{2}\right)^2\right)+\left(\sqrt{5}\right)^2-2\sqrt{1^2+\left\(\frac{1}{2} \right)^2}\sqrt{5}\cos\theta$

$\displaystyle \frac{9}{4}=\frac{5}{4}+5-5\cos\theta$

$\displaystyle 4=5\cos\theta$

$\displaystyle \theta=\cos^{-1}\left(\frac{4}{5}\right)\approx0.644$

Re: Solving trigonometric equations.

Hello MarkFL2

Quote:

Originally Posted by

**MarkFL2** To answer your question about extraneous roots, substitution into the original equation will in general show if the solution is extraneous or not.

I would have solved the equation much the same way you did:

$\displaystyle 2\cos\theta-\sin\theta=1$

$\displaystyle \sqrt{5}\sin\left(\theta+\pi-\tan^{-1}(2)\right)=1$

$\displaystyle \sin\left(\theta+\pi-\tan^{-1}(2)\right)=\frac{1}{\sqrt{5}}$

First solution:

$\displaystyle \theta+\pi-\tan^{-1}(2)=\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)$

$\displaystyle \theta=\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)+\tan^{-1}(2)-\pi$

Observing that in a $\displaystyle 1,2,\sqrt{5}$ right triangle, the two inverse trig. functions represent complementary angles, we have:

$\displaystyle \theta=\frac{\pi}{2}-\pi=-\frac{\pi}{2}$

Adding $\displaystyle 2\pi$ to get the angle in the desired range, we get:

$\displaystyle \theta=-\frac{\pi}{2}+2\pi=\frac{3\pi}{2}$

Second solution:

$\displaystyle \theta+\pi-\tan^{-1}(2)=\pi-\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)$

$\displaystyle \theta=\tan^{-1}(2)-\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)$

We could plug this into a calculator, but there is a slightly simpler representation.

Construct a right triangle with horizontal base 1, altitude 2 and hypotenuse $\displaystyle \sqrt{5}$ with the right angle at the lower right, then from the lower left vertex draw a line segment to the vertical leg at the point 1/2 unit above the right angle, and we have the angle subtended by this segment and the hypotenuse as $\displaystyle \theta$. Using the law of cosines, we find:

$\displaystyle \left(\frac{3}{2}\right)^2=\left(1^2+\left\(\frac{ 1}{2}\right)^2\right)+\left(\sqrt{5}\right)^2-2\sqrt{1^2+\left\(\frac{1}{2} \right)^2}\sqrt{5}\cos\theta$

$\displaystyle \frac{9}{4}=\frac{5}{4}+5-5\cos\theta$

$\displaystyle 4=5\cos\theta$

$\displaystyle \theta=\cos^{-1}\left(\frac{4}{5}\right)\approx0.644$

Thank you very much for taking the time to post that. Your method is similar in some respects. The one I used seems more basic, or at least I'm using it that way and I don't get an exact value for the first solution as you do. Using a right triangle I got $\displaystyle R = \sqrt{5}$ and $\displaystyle \tan\alpha = \dfrac{1}{2}$.

Which gave me $\displaystyle \sqrt{5}\cos(\theta + 0.464) = 1$ and then $\displaystyle \cos(\theta + 0.464) = \dfrac{1}{\sqrt{5}}$

I will try and work through your method.

2 Attachment(s)

Re: Solving trigonometric equations.

Attachment 23169

Notice we have:

$\displaystyle \alpha+\beta=\frac{\pi}{2}$

$\displaystyle \alpha=\tan^{-1}(2)$

$\displaystyle \beta=\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)$

thus:

$\displaystyle \sin^{-1}\left(\frac{1}{\sqrt{5}}\right)+\tan^{-1}(2)=\frac{\pi}{2}$

Attachment 23170

Notice that $\displaystyle \gamma=\alpha-\beta$

$\displaystyle \gamma=\tan^{-1}(2)-\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)$

Using the law of cosines allows you to find $\displaystyle \gamma$.

$\displaystyle \gamma=\cos^{-1}\left(\frac{4}{5}\right)=\sin^{-1}\left(\frac{3}{5}\right)=\tan^{-1}\left(\frac{3}{4}\right)$

Re: Solving trigonometric equations.

Hello MarkFL2(Bow)

Quote:

Originally Posted by

**MarkFL2** Attachment 23169
Notice we have:

$\displaystyle \alpha+\beta=\frac{\pi}{2}$

$\displaystyle \alpha=\tan^{-1}(2)$

$\displaystyle \beta=\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)$

thus:

$\displaystyle \sin^{-1}\left(\frac{1}{\sqrt{5}}\right)+\tan^{-1}(2)=\frac{\pi}{2}$

Attachment 23170
Notice that $\displaystyle \gamma=\alpha-\beta$

$\displaystyle \gamma=\tan^{-1}(2)-\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)$

Using the law of cosines allows you to find $\displaystyle \gamma$.

$\displaystyle \gamma=\cos^{-1}\left(\frac{4}{5}\right)=\sin^{-1}\left(\frac{3}{5}\right)=\tan^{-1}\left(\frac{3}{4}\right)$

It's extremely kind of you to go to the trouble to post that. It's very helpful, now I see where you were getting $\displaystyle \dfrac{\pi}{2}$, nice triangles. To be honest, at the moment I don't understand everything you've done, but I see that if I can it's going to make this topic very much more engaging. I'm going to look at it in more detail and get back to you if I have any questions. If that's alright.

Thanks again for all your help :)

Re: Solving trigonometric equations.

I just noticed a much simpler method to find $\displaystyle \gamma$. Rather than using the law of cosines, we may use the law of sines as well:

$\displaystyle \frac{\sin\gamma}{\frac{3}{2}}=\frac{\sin\beta}{ \frac{\sqrt{5}}{2}}$

Multiplying through by 3/2 and recalling $\displaystyle \sin\beta=\frac{1}{\sqrt{5}}$, we have:

$\displaystyle \sin\gamma=\frac{3}{5}$

$\displaystyle \gamma=\sin^{-1}\left(\frac{3}{5}\right)$

All of this really points out that the method used by **skeeter** is the way to go. (Cool)

Re: Solving trigonometric equations.

Hi MarkFL2

Quote:

Originally Posted by

**MarkFL2** I just noticed a much simpler method to find $\displaystyle \gamma$. Rather than using the law of cosines, we may use the law of sines as well:

$\displaystyle \frac{\sin\gamma}{\frac{3}{2}}=\frac{\sin\beta}{ \frac{\sqrt{5}}{2}}$

Multiplying through by 3/2 and recalling $\displaystyle \sin\beta=\frac{1}{\sqrt{5}}$, we have:

$\displaystyle \sin\gamma=\frac{3}{5}$

$\displaystyle \gamma=\sin^{-1}\left(\frac{3}{5}\right)$

All of this really points out that the method used by **skeeter** is the way to go. (Cool)

Thank you for that. I think you are right, the method used by skeeter is definately the way to go in this case and I used it to successfully solve other similiar question. However there are questions in which it states you have to solve equations of the form $\displaystyle {a}\cos\theta + {b}\sin\theta = {c}$ using the method which the book calls the 'harmonic form', of which your method seems to be an extension. Both make use of all three trigonometric ratios in a right angle triangle. Although I can use the method in the book I don't understand how it works. I think if I look at it more closely together with what you have shown me it's going to make a lot more sense and perhaps I can use it to find exact values, where they exist, as you did.

Thanks again for your efforts. :)