Danyal124
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I don’t understand how taking away angle gets the angle in between I understand use the sine graph to get the values but confused at the answerName:  DEADB15F-C96E-4102-802A-D000BCAA6927.jpeg
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0le
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Does this diagram from wiki help you understand?:
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Danyal124
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No I’ve never seen the unit circle in a level math could you try another way to explain this thank you
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mqb2766
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(Original post by Danyal124)
No I’ve never seen the unit circle in a level math could you try another way to explain this thank you
https://www.mathsisfun.com/geometry/unit-circle.html
Is decent. It's really just Pythagoras and basic trig.
Describing a circle as a parametric curve Is one of the simplest examples of parametric curves and hence worth understanding.

So the start/end of the arc are the start/end values of t. So subtract to get the angular width.
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Danyal124
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sorry i dont think the unit circle is making this clearer could you perphaps link this back to the equaton for angle in circle or an alternative route
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mqb2766
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(Original post by Danyal124)
sorry i dont think the unit circle is making this clearer could you perphaps link this back to the equaton for angle in circle or an alternative route
Why not pick some thetas and calculate the x,y coordinates and plot them (try and understand) on the curve. Understanding a unit circle in terms (cos(theta), sin(theta)) is fairly fundamental and underpins the sin^2+cos^2=1, parametric curves, sketching cos and sin curves, ...

At the end of the day, you have to understand how the arc is generated by the parametric curve, and the arc is part of the unit circle, so ..
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(Original post by Danyal124)
No I’ve never seen the unit circle in a level math could you try another way to explain this thank you
Each radius drawn on the circle is effectively the hypotenuse of a right-angled triangle. In the top right quadrant, the "adjacent" is a projection onto the positive x-axis and the "opposite" in the (projected) y-axis.

So for example, look at how 30^\circ is made by drawing a line which is 30^\circ to the x-axis. This line happens to be equal to 1 because it is also the radius of the circle. But using trigonometry, we can also make a right-angled triangle, setting the line as the hypotenuse, using the projection along the x-axis as the "adjacent" and the projection along the y-axis as the "opposite".

So from trig:
\cos \theta = \frac{adj}{hyp}
(hyp) \cos \theta = adj
(1) \cos \theta = adj
 \cos \theta = adj

This is shown in the figure below:
Image
Fig. 1. https://www.khanacademy.org/math/alg...-circle-review

Similarly for sine:
\sin \theta = \frac{opp}{hyp}
(hyp) \sin \theta = opp
(1) \sin \theta = opp
 \sin \theta = opp

Now you can vary the angle theta going from the positive x-axis anti-clockwise all the way back around again. A complete revolution is 360^\circ. In radians, 360^\circ = 2\pi. You can build the unit circle (a circle with radius of 1) by effectively making these triangles in each quadrant of the graph. But a pattern emerges because there is a symmtery between the triangles you construct. So you find that when you build a triangle in the top left corner that makes 30^\circ with the negative x-axis. But it revolves 150^\circ counter-clockwise from the positive x-axis and this leads to relationships such as  \sin (30^\circ) = \sin(150^\circ) = 0.5. You can see this more clear when you plot the sine function:

Image
https://www.mathsisfun.com/algebra/t...an-graphs.html

So finally, you are left with why do the two angles subtract. Well imagine all the angles start from the positive x axis and revolve counter-clockwise. They can go around the circle once (360^\circ) or twice ((2) (360^\circ) = 720^\circ ) and so fourth - but in this case it just begins to repeat again! If you have an angle in radians of \frac{3\pi}{4} you are in the top left quadrant. You can then figure out where the other angle is and work out what the angle is in between the two.

As an aside, you can therefore immediately derive the relationship:
\cos^2\theta + \sin^2\theta = 1 as follows:

a^2\ +b^2 = c^2 (Pythagoras theorem)
adj^2\ +opp^2 = hyp^2
(\cos \theta)^2 +(\sin \theta)^2 = (1)^2 (we showed this in Fig. 1)
(\cos \theta)(\cos \theta)+(\sin \theta)(\sin \theta)= 1
\cos^2\theta + \sin^2\theta = 1

Why is this all useful? Aside from the mathematical interest, well it often comes up in signal work and vibrations analysis. You often have use sine functions to model vibrations on surfaces or to model signals. You may process signals using Fourier analysis which uses this work etc. All things used in the real world to build engineering components or study nature.
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