# C2 - Trigonometry question involving radians

Hi there, so I am doing this question:

And I get the first solution to be 8pi/15. I got this by doing the inverse cos of 0.5 which gave me 1/3pi. I then added pi/5 to find the first solution.

However, I cant figure out how to find the second solution. The way I usually do it is via the methods of using (cos 360-ϴ), cos(360+ϴ), and cos(ϴ-360). I know that 2pi radians = 360 degrees, but I don't know how I can work this one out using that method, especially as it goes from -pi to pi (-180 to 180 degrees).

Can you please explain it to me? The mark scheme gets the second solution to be -2pi/15, but I can't see how you get that using the method I know, and also how you alter the method I know so that it fits within the -180 to 180 degree range.

Thanks
Original post by blobbybill
Hi there, so I am doing this question:

And I get the first solution to be 8pi/15. I got this by doing the inverse cos of 0.5 which gave me 1/3pi. I then added pi/5 to find the first solution.

However, I cant figure out how to find the second solution. The way I usually do it is via the methods of using (cos 360-ϴ), cos(360+ϴ), and cos(ϴ-360). I know that 2pi radians = 360 degrees, but I don't know how I can work this one out using that method, especially as it goes from -pi to pi (-180 to 180 degrees).

Can you please explain it to me? The mark scheme gets the second solution to be -2pi/15, but I can't see how you get that using the method I know, and also how you alter the method I know so that it fits within the -180 to 180 degree range.

Thanks

So you essentially want to solve $\cos x = \frac{1}{2}$. This has several solutions: $x = \arccos \frac{1}{2}, 2\pi + \arccos \frac{1}{2}, \arccos \frac{1}{2} - 2\pi$, etc...

In this case $x = \theta - \frac{\pi}{5}$. So $\theta - \frac{\pi}{5} = \arccos \frac{1}{2}, \arccos \frac{1}{2} - 2\pi, \arccos \frac{1}{2} + 2\pi$

then you add pi/5 to each of these solutions, throwing away any that's not in the required range
Original post by Zacken
So you essentially want to solve $\cos x = \frac{1}{2}$. This has several solutions: $x = \arccos \frac{1}{2}, 2\pi + \arccos \frac{1}{2}, \arccos \frac{1}{2} - 2\pi$, etc...

In this case $x = \theta - \frac{\pi}{5}$. So $\theta - \frac{\pi}{5} = \arccos \frac{1}{2}, \arccos \frac{1}{2} - 2\pi, \arccos \frac{1}{2} + 2\pi$

then you add pi/5 to each of these solutions, throwing away any that's not in the required range

What about cos(360-ϴ) as well? You didn't mention it, but its the other one that I Have. Would you try that one as well, add pi/5 to each and then only keep the ones in the range?

Thanks ever so much by the way, I really appreciate your help!
For a second, I thought this was GCSE Maths and wanted to cry since I had no idea what I was looking at. Luckily- it isn't Still not gonna get that Level 9/8 thou :/
Original post by blobbybill
What about cos(360-ϴ) as well? You didn't mention it, but its the other one that I Have. Would you try that one as well, add pi/5 to each and then only keep the ones in the range?

Thanks ever so much by the way, I really appreciate your help!

If you mean $\theta - \frac{\pi}{5} = 2\pi - \arccos \frac{1}{2}$, then yeah: add pi/5, see if it's still in the range, keep it if it is, throw away if it isn't
Original post by Zacken
If you mean $\theta - \frac{\pi}{5} = 2\pi - \arccos \frac{1}{2}$, then yeah: add pi/5, see if it's still in the range, keep it if it is, throw away if it isn't

I have tried the question, done what you said, and When I add pi/5 to all four solutions (ignoring the range for now), I don't get the solution of -2pi/15 that the mark scheme has for the second solution. Can you help me please? I have tried the method you said and I don't see where I have gone wrong.

Thanks
Original post by blobbybill
I have tried the question, done what you said, and When I add pi/5 to all four solutions (ignoring the range for now), I don't get the solution of -2pi/15 that the mark scheme has for the second solution. Can you help me please? I have tried the method you said and I don't see where I have gone wrong.

Thanks

@RDKGames I wonder if you can help me please. I found cos^-1(1/2) which gave me pi/3, I then got the four solutions by doing (2pi - theta), (2pi + theta), etc and then I added pi/5 to all of those solutions at the end, like RizK suggested I did (and how I got taught it).

However, I just cannot get the solution -2pi/15 like they get in the mark scheme. How do you get that? Where am I going wrong?

Thank you!
Original post by blobbybill
@RDKGames I wonder if you can help me please. I found cos^-1(1/2) which gave me pi/3, I then got the four solutions by doing (2pi - theta), (2pi + theta), etc and then I added pi/5 to all of those solutions at the end, like RizK suggested I did (and how I got taught it).

However, I just cannot get the solution -2pi/15 like they get in the mark scheme. How do you get that? Where am I going wrong?

Thank you!

Look back at your range, you are going way out of it with with $2\pi$

The only two solutions in that range are $\theta - \frac{\pi}{5}=\pm \frac{\pi}{3}$ now add the $\frac{\pi}{5}$ onto both.
(edited 7 years ago)
Original post by RDKGames
Look back at your range, you are going way out of with with $2\pi$

The only two solutions in that range are $\theta - \frac{\pi}{5}=\pm \frac{\pi}{3}$ now add the $\frac{\pi}{5}$ onto both.

Yeah, RizK told me to ignore the range until the end and then discard the solutions that aren't within the range at the very end.

How do you get the -pi/3? When I did inverse cos of 0.5, I got pi/3. Using the trig rules I got taught which are: cos theta = cos (360-theta) = cos(360+theta) = cos(theta-360), I don't see how you get -pi/3.
Original post by blobbybill
Yeah, RizK told me to ignore the range until the end and then discard the solutions that aren't within the range at the very end.

How do you get the -pi/3? When I did inverse cos of 0.5, I got pi/3. Using the trig rules I got taught which are: cos theta = cos (360-theta) = cos(360+theta) = cos(theta-360), I don't see how you get -pi/3.

Remember that $\cos(x)=\cos(-x)$, so whatever is inside cosine can be either sign. Thus, when you arccos both sides of $\cos(x-\frac{\pi}{5})=\frac{1}{2}$ then that means $x-\frac{\pi}{5}=\pm \arccos(\frac{1}{2})$

Then you can just keep adding/subtracting $2\pi$ for more solutions, obviously you dont need to here as it will get you out of your range, so you just leave it as it is and add on $\frac{\pi}{5}$
(edited 7 years ago)
Original post by RDKGames
Remember that $\cos(x)=\cos(-x)$, so whatever is inside cosine can be either sign. Thus, when you arccos both sides of $\cos(x-\frac{\pi}{5})=\frac{1}{2}$ then that means $x-\frac{\pi}{5}=\pm \arccos(\frac{1}{2})$

Then you can just keep adding/subtracting $2\pi$ for more solutions, obviously you dont need to here as it will get your out of your range, so you just leave it as it is and add on $\frac{\pi}{5}$

Okay, thank you. So ignoring the range, when I do the things with cosθ = cos (360-θ), can you also flip that answer of cos(360-θ) and make it negative too to get another answer?

Eg, if I have cos20, and I do cos(360-20) to find the second solution (which would be cos(340)), could I also then have cos(-340) as another answer, and then just add whatever it is, like how it was pi/5 here, to each one to give the actual solutions? Is that right, can you do that? I think you can because as you said, cosθ = cos(-θ)

Thanks
Original post by blobbybill
Okay, thank you. So ignoring the range, when I do the things with cosθ = cos (360-θ), can you also flip that answer of cos(360-θ) and make it negative too to get another answer?

Eg, if I have cos20, and I do cos(360-20) to find the second solution (which would be cos(340)), could I also then have cos(-340) as another answer, and then just add whatever it is, like how it was pi/5 here, to each one to give the actual solutions? Is that right, can you do that? I think you can because as you said, cosθ = cos(-θ)

Thanks

Yes you can, at least I think by what you're describing. Try solving $-1-2\cos(x+\frac{\pi}{4})=0$ in the range $-\pi < x \leq \pi$ with that idea and see if you get the right answers which should be $\frac{-11}{12}\pi,\frac{5}{12}\pi$
Original post by RDKGames
Yes you can, at least I think by what you're describing. Try solving $-1-2\cos(x+\frac{\pi}{4})=0$ in the range $-\pi < x \leq \pi$ with that idea and see if you get the right answers which should be $\frac{-11}{12}\pi,\frac{5}{12}\pi$

Sorry for the late reply. You know when you said that Cos(theta) = cos(-theta), is that the same for sine too? I'm doing this past question and I get how on the solution they got the -0.84 for cos (because cos(theta) = cos(-theta)), but how did they get x to be -pi as well for the sin part? Can you explain that please? I tried doing it using these identities like I normally do and I just can't get that answer of x= -pi.

Thanks

Original post by blobbybill
Sorry for the late reply. You know when you said that Cos(theta) = cos(-theta), is that the same for sine too? I'm doing this past question and I get how on the solution they got the -0.84 for cos (because cos(theta) = cos(-theta)), but how did they get x to be -pi as well for the sin part? Can you explain that please? I tried doing it using these identities like I normally do and I just can't get that answer of x= -pi.

No its not the same for sine.

sin(-x)=-sin(x)

Anyway, the board there says sin(x)=sin(180-x) so thats where -pi comes from. If x=0 is a solution then so is pi-0. As sine repats every 2pi, subtracting 2pi from pi-0 gives -pi
Original post by RDKGames
No its not the same for sine.

sin(-x)=-sin(x)

Anyway, the board there says sin(x)=sin(180-x) so thats where -pi comes from. If x=0 is a solution then so is pi-0. As sine repats every 2pi, subtracting 2pi from pi-0 gives -pi

Ok thanks.

I suppose it's also okay to think of it as taking pi from 0?

Also, if sine repeats every 2pi, how often does cos repeat?
If you substitute theta-pi/5 for x
Then x = pi/3 and -pi/3
Add pi/5 to both
You get 8pi/15 and -2pi/15
Original post by Beth17294
If you substitute theta-pi/5 for x
Then x = pi/3 and -pi/3
Add pi/5 to both
You get 8pi/15 and -2pi/15

Yes, but the last post on this thread was six years ago.