Maths -Exact trig values

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Sb2005
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How do you work out exact trig values using the ones you already know ,without a calculator ?
e.g- the exact trig value of cos 360°
Any videos regarding this would be helpful
Last edited by Sb2005; 3 months ago
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RDKGames
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(Original post by Sb2005)
How do you work out exact trig values using the ones you already know ,without a calculator ?
e.g- the exact trig value of cos 360°
Any videos regarding this would be helpful
Bunch of things to keep in mind.

Firstly, know your usual sines/cosines/tangents of typical angles 0,30,45,60,90,180,360.

Then, keep in mind periodicity of these functions. Sine & cosine repeat themselves every 360 degrees, so you can freely add/subtract multiples of 360 onto/from the argument ... so cos(420) = cos(420-360) = cos(60). Tangent repeats itself every 180 degrees so the same idea holds just with multiples of 180 instead.

Secondly, keep in mind some identities such as:

* sin(-x) = -sin(x) [ e.g. we can say that sin(-30) = -sin(30) ]
* cos(-x) = cos(x) [ e.g. we can say that cos(-45) = cos(45) ]

* sin(x) = sin(180-x) [ e.g. we can say that sin(120) = sin(180-120) = sin(60) ]
* cos(x) = cos(360-x) [ e.g. we can say that cos(315) = cos(360-315) = cos(45) ]

* tan(x) is ....(by definition) ... sin(x)/cos(x) ... so knowing the above sine and cosine tricks is sufficient to also adress tangent angles [ e.g. tan(-330) = sin(-330)/cos(-330) = sin(-330 + 360)/cos(-330 + 360) = sin(30)/cos(30) = tan(30)]

* sin(x) = cos(90-x) ... [ e.g. sin(120) = cos(90-120) = cos(-30) = cos(30) ]
* cos(x) = sin(90-x) ... these two are useful if you want to convert from sine to cosine and vice versa.
Last edited by RDKGames; 3 months ago
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ElMoro
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I'd like to suggest a different approach to remembering the graphs of \sin and \cos. Namely, the idea of the unit circle.

That is, consider a circle centred on the origin with a radius of 1:

Image

What's so important about the unit circle?

There is one very simple thing to remember about the unit circle which makes deriving trigonometric values and identities very easy.

That is: If you draw a line from the origin to a point on the circle such that the angle between the line you drew and the x-axis is t (anti-clockwise) then the point on the circle will have coordinates (\cos t, \sin t).

As the say, a picture's worth a thousand words:

Image


Let's see how this diagram can help us determine some common trigonometric values.

Let's say we want to work out \sin 90^{\circ} and \cos 90^{\circ}.

Let's draw a line at 90° to the x-axis:

Image

Clearly, the point (on the circle) we touch is (0, 1) so that means we can conclude that \cos 90^{\circ} = 0 and \sin 90^{\circ} = 1.

Now, let's say we want to work out \sin 0^{\circ} and \cos 0^{\circ}.

Let's draw a line at 0° to the x-axis (i.e. along the x-axis):

Image

In this case, the point (on the circle) we touch is (1, 0) so that means we can conclude that \cos 90^{\circ} = 1 and \sin 90^{\circ} = 0.

Have a go at determining \cos 180^{\circ}, \sin 180^{\circ}, \cos 270^{\circ}, \sin 270^{\circ} using the unit circle.

Eventually, you'll likely just be able to recall these values due to familiarity but I found the unit circle an invaluable tool to derive values to build that familiarity.

It's also useful for deriving trigonometric identities:
Spoiler:
Show
For example, suppose we have an angle \theta and want to determine which other angle "has the same \sin".

Well, we know that \sin \theta is given by the y-coordinate of a point on the circle. Suppose \theta is acute:
We can see that, by symmetry, there are at most two points on the circle which have the same y-coordinate:
Name:  Screenshot 2022-01-26 at 22.53.06.png
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The point at the end of the red line and the point at the end of the blue line. The anti-clockwise angles are \theta and 180^{\circ} - \theta.

This gives us the identity \sin \theta \equiv \sin (180^{\circ} - \theta ) (we assumed that /theta was acute for the diagram's sake but this identity holds more generally).

Have a go at:
(i) Deriving a similar identity for \cos \theta using symmetry
(ii) Determining a relationship between \sin \theta and \sin (360^{\circ} - \theta )
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Last edited by ElMoro; 3 months ago
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Muttley79
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(Original post by Sb2005)
How do you work out exact trig values using the ones you already know ,without a calculator ?
e.g- the exact trig value of cos 360°
Any videos regarding this would be helpful
This is a GCSE topic - post #2 is far too complex.

What have you been taught about this?


https://mathsmadeeasy.co.uk/gcse-mat...nd-worksheets/
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Sb2005
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Thank you very much to all of you!
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