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Plotting a function graph/ sin(2*t)

Graphing y = sin(2*t)

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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f(t) = sin(2*t)
f(t)=sin⁡(2t)f{\left(t \right)} = \sin{\left(2 t \right)} 
f = sin(2*t)
The graph of the function
-5.0-4.0-3.0-2.0-1.05.00.01.02.03.04.02-2
The points of intersection with the X-axis coordinate
Graph of the function intersects the axis T at f = 0
so we need to solve the equation:
sin⁡(2t)=0\sin{\left(2 t \right)} = 0
Solve this equation
The points of intersection with the axis T:

Analytical solution
t1=0t_{1} = 0
t2=π2t_{2} = \frac{\pi}{2}
Numerical solution
t1=−31.4159265358979t_{1} = -31.4159265358979
t2=51.8362787842316t_{2} = 51.8362787842316
t3=−36.1283155162826t_{3} = -36.1283155162826
t4=4.71238898038469t_{4} = 4.71238898038469
t5=89.5353906273091t_{5} = 89.5353906273091
t6=36.1283155162826t_{6} = 36.1283155162826
t7=−59.6902604182061t_{7} = -59.6902604182061
t8=−42.4115008234622t_{8} = -42.4115008234622
t9=21.9911485751286t_{9} = 21.9911485751286
t10=43.9822971502571t_{10} = 43.9822971502571
t11=15.707963267949t_{11} = 15.707963267949
t12=−53.4070751110265t_{12} = -53.4070751110265
t13=−80.1106126665397t_{13} = -80.1106126665397
t14=7.85398163397448t_{14} = 7.85398163397448
t15=−29.845130209103t_{15} = -29.845130209103
t16=−6.28318530717959t_{16} = -6.28318530717959
t17=78.5398163397448t_{17} = 78.5398163397448
t18=−23.5619449019235t_{18} = -23.5619449019235
t19=−17.2787595947439t_{19} = -17.2787595947439
t20=14.1371669411541t_{20} = 14.1371669411541
t21=58.1194640914112t_{21} = 58.1194640914112
t22=94.2477796076938t_{22} = 94.2477796076938
t23=12.5663706143592t_{23} = 12.5663706143592
t24=31.4159265358979t_{24} = 31.4159265358979
t25=−51.8362787842316t_{25} = -51.8362787842316
t26=−39.2699081698724t_{26} = -39.2699081698724
t27=80.1106126665397t_{27} = 80.1106126665397
t28=81.6814089933346t_{28} = 81.6814089933346
t29=−72.2566310325652t_{29} = -72.2566310325652
t30=64.4026493985908t_{30} = 64.4026493985908
t31=590.619418874881t_{31} = 590.619418874881
t32=67.5442420521806t_{32} = 67.5442420521806
t33=−20.4203522483337t_{33} = -20.4203522483337
t34=0t_{34} = 0
t35=−58.1194640914112t_{35} = -58.1194640914112
t36=−50.2654824574367t_{36} = -50.2654824574367
t37=−81.6814089933346t_{37} = -81.6814089933346
t38=86.3937979737193t_{38} = 86.3937979737193
t39=−64.4026493985908t_{39} = -64.4026493985908
t40=−483.805268652828t_{40} = -483.805268652828
t41=−14.1371669411541t_{41} = -14.1371669411541
t42=−48.6946861306418t_{42} = -48.6946861306418
t43=26.7035375555132t_{43} = 26.7035375555132
t44=23.5619449019235t_{44} = 23.5619449019235
t45=−87.9645943005142t_{45} = -87.9645943005142
t46=6.28318530717959t_{46} = 6.28318530717959
t47=−61.261056745001t_{47} = -61.261056745001
t48=−119.380520836412t_{48} = -119.380520836412
t49=−37.6991118430775t_{49} = -37.6991118430775
t50=−7.85398163397448t_{50} = -7.85398163397448
t51=34.5575191894877t_{51} = 34.5575191894877
t52=45.553093477052t_{52} = 45.553093477052
t53=65.9734457253857t_{53} = 65.9734457253857
t54=20.4203522483337t_{54} = 20.4203522483337
t55=−28.2743338823081t_{55} = -28.2743338823081
t56=−67.5442420521806t_{56} = -67.5442420521806
t57=72.2566310325652t_{57} = 72.2566310325652
t58=−43.9822971502571t_{58} = -43.9822971502571
t59=73.8274273593601t_{59} = 73.8274273593601
t60=−9.42477796076938t_{60} = -9.42477796076938
t61=−75.398223686155t_{61} = -75.398223686155
t62=113.097335529233t_{62} = 113.097335529233
t63=−95.8185759344887t_{63} = -95.8185759344887
t64=−97.3893722612836t_{64} = -97.3893722612836
t65=87.9645943005142t_{65} = 87.9645943005142
t66=42.4115008234622t_{66} = 42.4115008234622
t67=95.8185759344887t_{67} = 95.8185759344887
t68=59.6902604182061t_{68} = 59.6902604182061
t69=29.845130209103t_{69} = 29.845130209103
t70=100.530964914873t_{70} = 100.530964914873
t71=−73.8274273593601t_{71} = -73.8274273593601
t72=−40.8407044966673t_{72} = -40.8407044966673
t73=48.6946861306418t_{73} = 48.6946861306418
t74=28.2743338823081t_{74} = 28.2743338823081
t75=70.6858347057703t_{75} = 70.6858347057703
t76=−94.2477796076938t_{76} = -94.2477796076938
t77=−89.5353906273091t_{77} = -89.5353906273091
t78=−21.9911485751286t_{78} = -21.9911485751286
t79=56.5486677646163t_{79} = 56.5486677646163
t80=−65.9734457253857t_{80} = -65.9734457253857
t81=−15.707963267949t_{81} = -15.707963267949
t82=−45.553093477052t_{82} = -45.553093477052
t83=−1.5707963267949t_{83} = -1.5707963267949
t84=−83.2522053201295t_{84} = -83.2522053201295
t85=−86.3937979737193t_{85} = -86.3937979737193
t86=92.6769832808989t_{86} = 92.6769832808989
t87=50.2654824574367t_{87} = 50.2654824574367
t88=37.6991118430775t_{88} = 37.6991118430775
t89=1.5707963267949t_{89} = 1.5707963267949
The points of intersection with the Y axis coordinate
The graph crosses Y axis when t equals 0:
substitute t = 0 to sin(2*t).
sin⁡(0⋅2)\sin{\left(0 \cdot 2 \right)}
The result:
f(0)=0f{\left(0 \right)} = 0
The point:
(0, 0)
Extrema of the function
In order to find the extrema, we need to solve the equation
ddtf(t)=0\frac{d}{d t} f{\left(t \right)} = 0
(the derivative equals zero),
and the roots of this equation are the extrema of this function:
ddtf(t)=\frac{d}{d t} f{\left(t \right)} =
the first derivative
2cos⁡(2t)=02 \cos{\left(2 t \right)} = 0
Solve this equation
The roots of this equation
t1=π4t_{1} = \frac{\pi}{4}
t2=3π4t_{2} = \frac{3 \pi}{4}
The values of the extrema at the points:
 pi    
(--, 1)
 4     

 3*pi     
(----, -1)
  4


Intervals of increase and decrease of the function:
Let's find intervals where the function increases and decreases, as well as minima and maxima of the function, for this let's look how the function behaves itself in the extremas and at the slightest deviation from:
Minima of the function at points:
t1=3π4t_{1} = \frac{3 \pi}{4}
Maxima of the function at points:
t1=π4t_{1} = \frac{\pi}{4}
Decreasing at intervals
(−∞,π4]∪[3π4,∞)\left(-\infty, \frac{\pi}{4}\right] \cup \left[\frac{3 \pi}{4}, \infty\right)
Increasing at intervals
[π4,3π4]\left[\frac{\pi}{4}, \frac{3 \pi}{4}\right]
Inflection points
Let's find the inflection points, we'll need to solve the equation for this
d2dt2f(t)=0\frac{d^{2}}{d t^{2}} f{\left(t \right)} = 0
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
d2dt2f(t)=\frac{d^{2}}{d t^{2}} f{\left(t \right)} =
the second derivative
−4sin⁡(2t)=0- 4 \sin{\left(2 t \right)} = 0
Solve this equation
The roots of this equation
t1=0t_{1} = 0
t2=π2t_{2} = \frac{\pi}{2}

Сonvexity and concavity intervals:
Let’s find the intervals where the function is convex or concave, for this look at the behaviour of the function at the inflection points:
Concave at the intervals
(−∞,0]∪[π2,∞)\left(-\infty, 0\right] \cup \left[\frac{\pi}{2}, \infty\right)
Convex at the intervals
[0,π2]\left[0, \frac{\pi}{2}\right]
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at t->+oo and t->-oo
lim⁡t→−∞sin⁡(2t)=⟨−1,1⟩\lim_{t \to -\infty} \sin{\left(2 t \right)} = \left\langle -1, 1\right\rangle
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=⟨−1,1⟩y = \left\langle -1, 1\right\rangle
lim⁡t→∞sin⁡(2t)=⟨−1,1⟩\lim_{t \to \infty} \sin{\left(2 t \right)} = \left\langle -1, 1\right\rangle
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=⟨−1,1⟩y = \left\langle -1, 1\right\rangle
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of sin(2*t), divided by t at t->+oo and t ->-oo
lim⁡t→−∞(sin⁡(2t)t)=0\lim_{t \to -\infty}\left(\frac{\sin{\left(2 t \right)}}{t}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡t→∞(sin⁡(2t)t)=0\lim_{t \to \infty}\left(\frac{\sin{\left(2 t \right)}}{t}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
Even and odd functions
Let's check, whether the function even or odd by using relations f = f(-t) и f = -f(-t).
So, check:
sin⁡(2t)=−sin⁡(2t)\sin{\left(2 t \right)} = - \sin{\left(2 t \right)}
- No
sin⁡(2t)=sin⁡(2t)\sin{\left(2 t \right)} = \sin{\left(2 t \right)}
- Yes
so, the function
is
odd
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