# Math Help - Solving trigonometric equations.

1. ## Solving trigonometric equations.

Hello

The question was:

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

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

I was able to solve this using the harmonic form, expressing the $LHS$ in terms of $R\cos(\theta + \alpha)$. 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.

2. ## Re: Solving trigonometric equations.

Originally Posted by Furyan
Hello

The question was:

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

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

I was able to solve this using the harmonic form, expressing the $LHS$ in terms of $R\cos(\theta + \alpha)$. 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.
$2\cos{t} = \sin{t} + 1$

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

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

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

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

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

3. ## 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.

4. ## Re: Solving trigonometric equations.

Hello Skeeter

Originally Posted by skeeter
$2\cos{t} = \sin{t} + 1$

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

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

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

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

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 $\sin t$. 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 $\sin t = \dfrac{3}{5}$ gives me $t = 0.644$ and $t = 2.50$ and $\sin t = -1$ gives me $t = 4.71 = (\dfrac{3}{2}\pi)$. The first and the last of these solutions I got using the harmonic form, which, since you mentioned them, leads me to believe that $t = 2.50$ is an extraneous root, but looking at the graph of $y = \sin x$ 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 $t = 2.50$ is not a solution. Is that how I check for extraneous roots, by seeing if they work?

5. ## Re: Solving trigonometric equations.

Hello ILikeSerena

Originally Posted by ILikeSerena
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 it what you meant with the harmonic form.
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.

6. ## 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:

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

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

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

First solution:

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

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

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

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

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

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

Second solution:

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

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

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 $\sqrt{5}$ 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 $\theta$. Using the law of cosines, we find:

$\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$

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

$4=5\cos\theta$

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

7. ## Re: Solving trigonometric equations.

Hello MarkFL2

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:

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

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

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

First solution:

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

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

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

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

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

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

Second solution:

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

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

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 $\sqrt{5}$ 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 $\theta$. Using the law of cosines, we find:

$\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$

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

$4=5\cos\theta$

$\theta=\cos^{-1}\left(\frac{4}{5}\right)\approx0.644$
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 $R = \sqrt{5}$ and $\tan\alpha = \dfrac{1}{2}$.

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

I will try and work through your method.

8. ## Re: Solving trigonometric equations.

Notice we have:

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

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

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

thus:

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

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

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

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

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

9. ## Re: Solving trigonometric equations.

Hello MarkFL2

Originally Posted by MarkFL2

Notice we have:

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

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

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

thus:

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

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

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

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

$\gamma=\cos^{-1}\left(\frac{4}{5}\right)=\sin^{-1}\left(\frac{3}{5}\right)=\tan^{-1}\left(\frac{3}{4}\right)$
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 $\dfrac{\pi}{2}$, 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

10. ## Re: Solving trigonometric equations.

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

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

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

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

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

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

11. ## Re: Solving trigonometric equations.

Hi MarkFL2

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

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

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

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

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

All of this really points out that the method used by skeeter is the way to go.
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 ${a}\cos\theta + {b}\sin\theta = {c}$ 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.