Self teaching FP1; finding other solution in given interval

I actually forgot this

Anyway, say theta=cos^-1 (1/2) =60 degrees

For interval -180<theta<180 the other solution is -60 degrees.. How and why again? I forgot..

If it was say sin^-1 (1/2)= 30 degrees what would the other solution be in the same interval?

And finally, in radians, x=sin^-1 (1/(sqrt2))= PI/4. Why is the other solution 3PI/4 in the interval -PI<theta<PI?

Thanks in advance all

Do you know what a CAST diagram is? I think you should have done this in C2
Or you could draw a sin/cos graph to help you.

For the cos^-1 (1/2)= 60, you could draw a cosine graph and from 1/2 on the y-axis draw a straight line, and where the line touches the graph are your solutions. The first solution is 60 that you got from your calculator so you should be able to find the others in the given range using this method.

During 0 to 90 degrees COS is positive, 90 to 180 All are positive, 180 to 270 SIN is positive 270-360 TAN is positive, 360-450 COS is positive and so on etc?

Is there a quicker way to find these apart from drawing graphs?

During 0 to 90 degrees COS is positive, 90 to 180 All are positive, 180 to 270 SIN is positive 270-360 TAN is positive, 360-450 COS is positive and so on etc?

Is there a quicker way to find these apart from drawing graphs?

You don't need to know this CAST rubbish or draw graphs.

$\cos x = \cos (-x)$

$\cos x = \cos (x \pm 360)$

$\sin x = \sin (180-x)$

$\sin x = \sin (x \pm 360)$

$\tan x = \tan(x \pm 180)$

Exactly what I was looking for. Cheers Mr M

Although he did miss cos(x)=360-x

I actually forgot this

Anyway, say theta=cos^-1 (1/2) =60 degrees

For interval -180<theta<180 the other solution is -60 degrees.. How and why again? I forgot..

If it was say sin^-1 (1/2)= 30 degrees what would the other solution be in the same interval?

And finally, in radians, x=sin^-1 (1/(sqrt2))= PI/4. Why is the other solution 3PI/4 in the interval -PI<theta<PI?

Thanks in advance all

Simple way
For Sin, if you have Sin(x), the other solution is Sin(180-X) so if X = 20, it'll also = 160

For Cos, if you have Cos(X), the other solution is Cos(360-X) so if you have X = 20, it'll also = 340

And you know Sin and Cos cycle every 360° , if the limits are below 0 or above 360 you just subtract or add 360 to your two solutions!

So if limit is -360 to 720
For Sin30 it'll also = Sin(180-30) and then Subtract or add 360 to get the rest of the solutions till you hit your limit!

Inbox me if you need more help!

Hope that helped!

Posted from TSR Mobile

Of course.


For example, if x = 60 degrees, then x = 300 degrees. For proof, see graph .

Oh dear

He meant - are you sure that he missed it

I did mean that but I was also questioning the validity of his statement that cos x = 360 - x. Double whammy.

I did mean that but I was also questioning the validity of his statement that cos x = 360 - x. Double whammy.

Well, you did miss it. And yes, cos(x) = 360 - x.

Ah, well sorry for that .

Well, you did miss it. And yes, cos(x) = 360 - x. This is one of the first laws you learn, during C2. Here's the proof:
Cos(360-x) = Cos(360)Cos(x) + Sin(360)Sin(x)
Cos(360-x) = (1)Cos(x) + (0)Sin(x)
Cos(360-x) = Cos(x)

Plus, like I said, you can easily see it using the Cos(x) graph.

You don't need to know this CAST rubbish or draw graphs.

$\cos x = \cos (-x)$

$\cos x = \cos (x \pm 360)$

$\sin x = \sin (180-x)$

$\sin x = \sin (x \pm 360)$

$\tan x = \tan(x \pm 180)$

This is all that was posted, there is clearly no formula for Cos(x)=Cos(360-x) (Which is what I meant when I said Cos(x)=360-x, sorry for not typing that up correctly).

You don't need to know this CAST rubbish or draw graphs.

$\cos x = \cos (-x)$

$\cos x = \cos (x \pm 360)$ (LOOK AT ME)

$\sin x = \sin (180-x)$

$\sin x = \sin (x \pm 360)$

$\tan x = \tan(x \pm 180)$