For the first one.
=(2\sin^{2}(t)-1)\underset{\sim}{i}-\sin(2t)\underset{\sim}{j})
,
Now the "speed" of the particle will be the magnitude of the velocity, that is
|)
So we have
|= & \sqrt{\left(2\sin^{2}(t)-1\right)^{2}+\left(-\sin(2t)\right)^{2}}<br />\\ & =\sqrt{4\sin^{4}(t)-4\sin^{2}(t)+1+\sin^{2}(2t)}<br />\\ & =\sqrt{4\sin^{4}(t)-4\sin^{2}(t)+1+\left(2\sin(t)\cos(t)\right)^{2}}\:\:\:\:\left(\mathrm{as\:}\sin(2t)=2\sin(t)\cos(t)\right)<br />\\ & =\sqrt{4\sin^{4}(t)-4\sin^{2}(t)+1+4\sin^{2}(t)\cos^{2}(t)}<br />\\ & =\sqrt{4\sin^{2}(t)\left(\sin^{2}(t)+\cos^{2}(t)\right)+1-4\sin^{2}(t)}<br />\\ & =\sqrt{4\sin^{2}(t)\left(1\right)+1-4\sin^{2}(t)}\:\:\:\:\left(\mathrm{as\:}\sin^{2}(t)+\cos^{2}(t)=1)<br />\\ & =\sqrt{1}<br />\\ & =1<br />\end{alignedat})
The speed of the particle is constant.
EDIT: Not too sure about this one so hope this works.
Since we have three vectors that when added together result in 0, we know that we have a triangle formed. When the acute angles made between the vectors is added together, they should add to
)
.
Now we know our dot product
=\frac{\underset{\sim}{a}.\underset{\sim}{b}}{|\underset{\sim}{a}||\underset{\sim}{b}|})
, i.e.
)
Now lets let
- the angle between
and
be
- the angle between
and
be
- the angle between
and
be
Now
(since they are unit vectors)
So that leaves
)
)
)
So that leads to
+\cos(\beta)+\cos(\gamma))
.....[3]
Now since the length of all three sides is equal (as they are all unit vectors, so we have an equilateral triangle),
if we move the vectors so that they are tail-to-tail for the dot product(see attached diagram) then we will end up with an angle of
)
between then all,
will all be
.
+\cos(\frac{2\pi}{3})+\cos(\frac{2\pi}{3})<br />\\ & =-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}<br />\\ & =-\frac{3}{2}<br />\end{alignedat})
Hope that last one is right
EDIT: Made a small fix, attached diagram.
NOTE: The angle between the vectors in the first triangle is 
, )
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