What wrong am I doing while writing these parametric coordinates?

Question

I kinda understand that it should be $-60^\circ $ in the second picture. But I am getting that after looking at the result. I can't exactly understand what wrong I am doing.

PS: The length of the line is $1$ unit.

Answer 1

See that there are rules defined and followed conventionally The angles are always measured in anticlockwise sense from positive $X$ axis And so there are trigonometric rules defined based on these.

From the above animation, notice the sign of values of sin and cos in different quadrants $ $

All ratios are positive in 1st quadrant $ $

only sin and cosec are positive in 2nd quadrant $ $

only tan and cot are positive in 3rd quadrant as both cos and sin are negative

only cos and sec are positive in 2nd quadrant$ $

Like $cos(180 - \theta ) = - cos(\theta)$ (1)

$sin(180 - \theta ) = sin(\theta)$ (2) and many more

To your question the parametric coordinates should be Generalized one : $\Bigl( a+cos(\theta) , sin(\theta) \Bigr)$

At the instant of your picture i.e at an angle 120° it's coordinates would be:

• $\Bigl( a+cos(120) , sin(120) \Bigr)$
• $\Bigl( a - cos(60) , sin(60) \Bigr)$ as cos(120)= - cos(60) and sin(120) = sin(60) from (1) and (2)

Answer 2

It is really simple, you have point $(a,0)$ and want to move in a certain direction. For that you consider the vector $v=(x,y)$.

But now, $(x,y)=(\cos(\theta),\sin(\theta))$ works only when using the standard way to measure angles (from the $x$-axis counterclockwise). Do note that your new way to measure angles is just to reflect on the $y$-axis. So now, the correct way to get coordinates with that way of measuring angles is to take into account that reflection. $$(-x,y)= (\cos(\theta),\sin(\theta)) \iff (x,y)= (-\cos(\theta),\sin(\theta))$$

Also note that what you said about replacing $60$ by $-60$ wouldn't work due to the sine function.

Answer 3

I'll assume throughout the trigonometric identities$$\cos\theta=-\cos(180^\circ\pm\theta),\,\sin\theta=\sin(180^\circ-\theta)=-\sin(\theta-180^\circ).$$Whether you think of these identities as implying everything works out, or think of the diagrams as providing proof of these identities, is up to you.

In the first diagram, to $\binom{a}{0}$ you add the unit vector $\binom{\cos120^\circ}{\sin120^\circ}$. This second vector actually has negative $x$-coordinate, so the new $x$-coordinate is less than $a$ by $-\cos120^\circ=\cos60^\circ$.

In the second diagram, you don't add $\binom{\cos60^\circ}{\sin60^\circ}$, because the $60^\circ$ angle doesn't go anticlockwise from the portion of the $x$-axis right of $\binom{a}{0}$ (which is the usual approach, as used in the first picture); it goes clockwise from the portion of the $x$-axis left of $\binom{a}{0}$. So let's work out what happens this time.

The right-angled triangle this $60^\circ$ angle creates has base $\cos60^\circ=-\cos120^\circ$, but since it takes us left of $\binom{a}{0}$, the $x$-coordinate is reduced by $\cos60^\circ$, i.e. $\cos120^\circ$ is added to it. The triangle's opposite, $\sin60^\circ$, runs up rather than down, so the $y$-coordinate is actually increased, by $\sin60^\circ=\sin120^\circ$.

I'll close by addressing this part

it should be $-60^\circ$ in the second picture

Unsurprisingly, the change in position is the same by both methods, but let's rewrite it:$$\binom{\cos120^\circ}{\sin120^\circ}=-\binom{\cos(-60^\circ)}{\sin(-60^\circ)}.$$So you were right: in a way, $-60^\circ$ is a rotation angle we can think in terms of here. But why is there an overall $-$ sign in front of the position change, too?

Suppose a second unit vector anchored at $\binom{a}{0}$ had been "rotated $-60^\circ$ anticlockwise", i.e. rotated $60^\circ$ clockwise, to relocate an end initially at $\binom{a+1}{0}$ to $\binom{a}{0}+\binom{\cos(-60^\circ)}{\sin(-60^\circ)}.$. Said end would have ended up as the reflection in $\binom{a}{0}$ of the endpoint whose position we've been computing throughout, because a $120^\circ+60^\circ=180^\circ$ angle, or straight line, would have separated the two unit vectors. Therefore, the endpoint in your two diagrams is instead $\binom{a}{0}+\binom{\cos(-60^\circ)}{\sin(-60^\circ)}$.

Answer 4

What you are parameterizing is the circle of radius $1$ centered at $(a,0)$, counter-clockwise. That is given by $(a,0) + (\cos(t), \sin(t))$. For $t = 120º$ (it's equivalent in radians, anyway), you get what you are showing in your first picture.

But, if you put $t = 60º$, this corresponds to an angle of $60º$ measured counter-clockwise. That is, it should form an angle which is a half of the first angle you pictured.

To get that point in your second picture for an angle of $60º$, you must use another parametrization (a clockwise one, starting at $(a,0) - (1,0)$). That can be given by $(a,0) + (-\cos(t), \sin(t))$