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  • Graphing y =:
  • x/((x-1)*(x-4))
  • x(x-1)^3
  • x+(lnx)/x
  • x*cbrt((x-1)^2)

  • Identical expressions

  • sin^ two (t)/(two +sin(t))
  • sinus of squared (t) divide by (2 plus sinus of (t))
  • sinus of to the power of two (t) divide by (two plus sinus of (t))
  • sin2(t)/(2+sin(t))
  • sin2t/2+sint
  • sin²(t)/(2+sin(t))
  • sin to the power of 2(t)/(2+sin(t))
  • sin^2t/2+sint
  • sin^2(t) divide by (2+sin(t))

  • Similar expressions

  • sin^2(t)/(2-sin(t))
Plotting a function graph/ sin^2(t)/(2+sin(t))

Graphing y = sin^2(t)/(2+sin(t))

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

You have entered [src] 
           2     
        sin (t)  
f(t) = ----------
       2 + sin(t)
f(t)=sin⁡2(t)sin⁡(t)+2f{\left(t \right)} = \frac{\sin^{2}{\left(t \right)}}{\sin{\left(t \right)} + 2} 
f = sin(t)^2/(sin(t) + 2)
The graph of the function
-1.0-0.8-0.6-0.4-0.21.00.00.20.40.60.80.01.0
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⁡2(t)sin⁡(t)+2=0\frac{\sin^{2}{\left(t \right)}}{\sin{\left(t \right)} + 2} = 0
Solve this equation
The points of intersection with the axis T:

Analytical solution
t1=0t_{1} = 0
t2=πt_{2} = \pi
Numerical solution
t1=−56.5486675835621t_{1} = -56.5486675835621
t2=3.14159312185834t_{2} = 3.14159312185834
t3=−40.8407040289876t_{3} = -40.8407040289876
t4=53.4070757175476t_{4} = 53.4070757175476
t5=25.1327407744412t_{5} = 25.1327407744412
t6=−50.2654823212533t_{6} = -50.2654823212533
t7=−62.8318529476859t_{7} = -62.8318529476859
t8=−37.6991126284565t_{8} = -37.6991126284565
t9=12.5663704137215t_{9} = 12.5663704137215
t10=−97.3893724084397t_{10} = -97.3893724084397
t11=−59.6902604559996t_{11} = -59.6902604559996
t12=−12.566370432922t_{12} = -12.566370432922
t13=78.5398162107716t_{13} = 78.5398162107716
t14=31.4159267173782t_{14} = 31.4159267173782
t15=−84.8230011807336t_{15} = -84.8230011807336
t16=43.982297991191t_{16} = 43.982297991191
t17=−18.8495563752342t_{17} = -18.8495563752342
t18=−75.3982239084087t_{18} = -75.3982239084087
t19=−62.8318535236677t_{19} = -62.8318535236677
t20=31.4159264181377t_{20} = 31.4159264181377
t21=−21.9911485863129t_{21} = -21.9911485863129
t22=−81.6814090403179t_{22} = -81.6814090403179
t23=75.3982240661504t_{23} = 75.3982240661504
t24=−31.4159267473569t_{24} = -31.4159267473569
t25=69.1150379252292t_{25} = 69.1150379252292
t26=15.7079634846579t_{26} = 15.7079634846579
t27=−69.1150388783611t_{27} = -69.1150388783611
t28=−91.1061871295728t_{28} = -91.1061871295728
t29=−3.14159245222419t_{29} = -3.14159245222419
t30=87.9645943344402t_{30} = 87.9645943344402
t31=100.530964735934t_{31} = 100.530964735934
t32=−69.1150387611399t_{32} = -69.1150387611399
t33=25.1327413267662t_{33} = 25.1327413267662
t34=47.1238897117841t_{34} = 47.1238897117841
t35=75.3982238673193t_{35} = 75.3982238673193
t36=6.28318528365976t_{36} = 6.28318528365976
t37=−47.1238895535581t_{37} = -47.1238895535581
t38=18.849555556969t_{38} = 18.849555556969
t39=53.4070755016345t_{39} = 53.4070755016345
t40=−65.9734457633727t_{40} = -65.9734457633727
t41=21.9911485853088t_{41} = 21.9911485853088
t42=72.2566310277446t_{42} = 72.2566310277446
t43=−43.9822971752436t_{43} = -43.9822971752436
t44=75.3982235055873t_{44} = 75.3982235055873
t45=−15.7079632956568t_{45} = -15.7079632956568
t46=−72.2566308384363t_{46} = -72.2566308384363
t47=−84.8230016739514t_{47} = -84.8230016739514
t48=−40.8407045494855t_{48} = -40.8407045494855
t49=94.2477796093495t_{49} = 94.2477796093495
t50=−34.5575188166549t_{50} = -34.5575188166549
t51=40.8407043200489t_{51} = 40.8407043200489
t52=−94.2477794766245t_{52} = -94.2477794766245
t53=−91.1061866630345t_{53} = -91.1061866630345
t54=50.265482446211t_{54} = 50.265482446211
t55=59.6902606460951t_{55} = 59.6902606460951
t56=−53.4070752540144t_{56} = -53.4070752540144
t57=−3.14159283024245t_{57} = -3.14159283024245
t58=−87.9645943627279t_{58} = -87.9645943627279
t59=−81.6814088716349t_{59} = -81.6814088716349
t60=−12.5663706569192t_{60} = -12.5663706569192
t61=43.9822971690315t_{61} = 43.9822971690315
t62=91.1061874205003t_{62} = 91.1061874205003
t63=97.3893726603459t_{63} = 97.3893726603459
t64=−6.28318516617079t_{64} = -6.28318516617079
t65=−78.5398159761873t_{65} = -78.5398159761873
t66=84.8230017192229t_{66} = 84.8230017192229
t67=28.274333865516t_{67} = 28.274333865516
t68=81.6814091431634t_{68} = 81.6814091431634
t69=0t_{69} = 0
t70=65.9734457537561t_{70} = 65.9734457537561
t71=−25.1327416022221t_{71} = -25.1327416022221
t72=−56.5486677178641t_{72} = -56.5486677178641
t73=−37.699111878477t_{73} = -37.699111878477
t74=56.54866757482t_{74} = 56.54866757482
t75=−9.42477809933772t_{75} = -9.42477809933772
t76=34.5575190555905t_{76} = 34.5575190555905
t77=−100.530964734526t_{77} = -100.530964734526
t78=3.14159257938259t_{78} = 3.14159257938259
t79=9.42477834262643t_{79} = 9.42477834262643
t80=−65.9734468967977t_{80} = -65.9734468967977
t81=62.8318527164084t_{81} = 62.8318527164084
t82=−31.4159273869072t_{82} = -31.4159273869072
t83=37.6991119886324t_{83} = 37.6991119886324
t84=−47.1238899801267t_{84} = -47.1238899801267
t85=84.8230014707211t_{85} = 84.8230014707211
t86=18.8495572418557t_{86} = 18.8495572418557
t87=91.1061868470564t_{87} = 91.1061868470564
t88=−43.982298747563t_{88} = -43.982298747563
t89=−28.2743336768524t_{89} = -28.2743336768524
t90=−18.8495558093765t_{90} = -18.8495558093765
t91=47.1238902717198t_{91} = 47.1238902717198
t92=40.840704639321t_{92} = 40.840704639321
t93=103.672556339194t_{93} = 103.672556339194
t94=69.1150384602719t_{94} = 69.1150384602719
The points of intersection with the Y axis coordinate
The graph crosses Y axis when t equals 0:
substitute t = 0 to sin(t)^2/(2 + sin(t)).
sin⁡2(0)sin⁡(0)+2\frac{\sin^{2}{\left(0 \right)}}{\sin{\left(0 \right)} + 2}
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
2sin⁡(t)cos⁡(t)sin⁡(t)+2−sin⁡2(t)cos⁡(t)(sin⁡(t)+2)2=0\frac{2 \sin{\left(t \right)} \cos{\left(t \right)}}{\sin{\left(t \right)} + 2} - \frac{\sin^{2}{\left(t \right)} \cos{\left(t \right)}}{\left(\sin{\left(t \right)} + 2\right)^{2}} = 0
Solve this equation
The roots of this equation
t1=0t_{1} = 0
t2=−π2t_{2} = - \frac{\pi}{2}
t3=π2t_{3} = \frac{\pi}{2}
The values of the extrema at the points:
(0, 0)

 -pi     
(----, 1)
  2      

 pi      
(--, 1/3)
 2


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=0t_{1} = 0
Maxima of the function at points:
t1=−π2t_{1} = - \frac{\pi}{2}
t1=π2t_{1} = \frac{\pi}{2}
Decreasing at intervals
(−∞,−π2]∪[0,∞)\left(-\infty, - \frac{\pi}{2}\right] \cup \left[0, \infty\right)
Increasing at intervals
(−∞,0]∪[π2,∞)\left(-\infty, 0\right] \cup \left[\frac{\pi}{2}, \infty\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
−2sin⁡2(t)+2cos⁡2(t)+(sin⁡(t)+2cos⁡2(t)sin⁡(t)+2)sin⁡2(t)sin⁡(t)+2−4sin⁡(t)cos⁡2(t)sin⁡(t)+2sin⁡(t)+2=0\frac{- 2 \sin^{2}{\left(t \right)} + 2 \cos^{2}{\left(t \right)} + \frac{\left(\sin{\left(t \right)} + \frac{2 \cos^{2}{\left(t \right)}}{\sin{\left(t \right)} + 2}\right) \sin^{2}{\left(t \right)}}{\sin{\left(t \right)} + 2} - \frac{4 \sin{\left(t \right)} \cos^{2}{\left(t \right)}}{\sin{\left(t \right)} + 2}}{\sin{\left(t \right)} + 2} = 0
Solve this equation
The roots of this equation
t1=2atan⁡(CRootOf⁡(x8−4x6−6x5−12x4−6x3−4x2+1,0))t_{1} = 2 \operatorname{atan}{\left(\operatorname{CRootOf} {\left(x^{8} - 4 x^{6} - 6 x^{5} - 12 x^{4} - 6 x^{3} - 4 x^{2} + 1, 0\right)} \right)}
t2=2atan⁡(CRootOf⁡(x8−4x6−6x5−12x4−6x3−4x2+1,1))t_{2} = 2 \operatorname{atan}{\left(\operatorname{CRootOf} {\left(x^{8} - 4 x^{6} - 6 x^{5} - 12 x^{4} - 6 x^{3} - 4 x^{2} + 1, 1\right)} \right)}
t3=2atan⁡(CRootOf⁡(x8−4x6−6x5−12x4−6x3−4x2+1,2))t_{3} = 2 \operatorname{atan}{\left(\operatorname{CRootOf} {\left(x^{8} - 4 x^{6} - 6 x^{5} - 12 x^{4} - 6 x^{3} - 4 x^{2} + 1, 2\right)} \right)}
t4=2atan⁡(CRootOf⁡(x8−4x6−6x5−12x4−6x3−4x2+1,3))t_{4} = 2 \operatorname{atan}{\left(\operatorname{CRootOf} {\left(x^{8} - 4 x^{6} - 6 x^{5} - 12 x^{4} - 6 x^{3} - 4 x^{2} + 1, 3\right)} \right)}

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
[2atan⁡(CRootOf⁡(x8−4x6−6x5−12x4−6x3−4x2+1,3)),∞)\left[2 \operatorname{atan}{\left(\operatorname{CRootOf} {\left(x^{8} - 4 x^{6} - 6 x^{5} - 12 x^{4} - 6 x^{3} - 4 x^{2} + 1, 3\right)} \right)}, \infty\right)
Convex at the intervals
(−∞,2atan⁡(CRootOf⁡(x8−4x6−6x5−12x4−6x3−4x2+1,1))]∪[2atan⁡(CRootOf⁡(x8−4x6−6x5−12x4−6x3−4x2+1,2)),2atan⁡(CRootOf⁡(x8−4x6−6x5−12x4−6x3−4x2+1,3))]\left(-\infty, 2 \operatorname{atan}{\left(\operatorname{CRootOf} {\left(x^{8} - 4 x^{6} - 6 x^{5} - 12 x^{4} - 6 x^{3} - 4 x^{2} + 1, 1\right)} \right)}\right] \cup \left[2 \operatorname{atan}{\left(\operatorname{CRootOf} {\left(x^{8} - 4 x^{6} - 6 x^{5} - 12 x^{4} - 6 x^{3} - 4 x^{2} + 1, 2\right)} \right)}, 2 \operatorname{atan}{\left(\operatorname{CRootOf} {\left(x^{8} - 4 x^{6} - 6 x^{5} - 12 x^{4} - 6 x^{3} - 4 x^{2} + 1, 3\right)} \right)}\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⁡2(t)sin⁡(t)+2)=⟨0,1⟩\lim_{t \to -\infty}\left(\frac{\sin^{2}{\left(t \right)}}{\sin{\left(t \right)} + 2}\right) = \left\langle 0, 1\right\rangle
Let's take the limit
so,
equation of the horizontal asymptote on the left:
y=⟨0,1⟩y = \left\langle 0, 1\right\rangle
lim⁡t→∞(sin⁡2(t)sin⁡(t)+2)=⟨0,1⟩\lim_{t \to \infty}\left(\frac{\sin^{2}{\left(t \right)}}{\sin{\left(t \right)} + 2}\right) = \left\langle 0, 1\right\rangle
Let's take the limit
so,
equation of the horizontal asymptote on the right:
y=⟨0,1⟩y = \left\langle 0, 1\right\rangle
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of sin(t)^2/(2 + sin(t)), divided by t at t->+oo and t ->-oo
lim⁡t→−∞(sin⁡2(t)t(sin⁡(t)+2))=0\lim_{t \to -\infty}\left(\frac{\sin^{2}{\left(t \right)}}{t \left(\sin{\left(t \right)} + 2\right)}\right) = 0
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
lim⁡t→∞(sin⁡2(t)t(sin⁡(t)+2))=0\lim_{t \to \infty}\left(\frac{\sin^{2}{\left(t \right)}}{t \left(\sin{\left(t \right)} + 2\right)}\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⁡2(t)sin⁡(t)+2=sin⁡2(t)2−sin⁡(t)\frac{\sin^{2}{\left(t \right)}}{\sin{\left(t \right)} + 2} = \frac{\sin^{2}{\left(t \right)}}{2 - \sin{\left(t \right)}}
- No
sin⁡2(t)sin⁡(t)+2=−sin⁡2(t)2−sin⁡(t)\frac{\sin^{2}{\left(t \right)}}{\sin{\left(t \right)} + 2} = - \frac{\sin^{2}{\left(t \right)}}{2 - \sin{\left(t \right)}}
- No
so, the function
not is
neither even, nor odd
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