Tan (Ф/2)=⁺₋√[(1-cosФ)/(1+cosФ)]
if π<Ф<3π/2;
then, Where is Ф/2??
π/2<Ф/2<3π/4; therefore Ф/2 is in the second quadrant; then tan (Ф/2) will have a negative value.
tan(Ф/2)=-√[(1-cosФ)/(1+cosФ)]
Now, we have to find the value of cos Ф.
tan (Ф)=4/3
1+tan²Ф=sec²Ф
1+(4/3)²=sec²Ф
sec²Ф=1+16/9
sec²Ф=(9+16)/9
sec²Ф=25/9
sec Ф=-√(25/9) (sec²Ф will have a negative value, because Ф is in the sec Ф=-5/3 third quadrant).
cos Ф=1/sec Ф
cos Ф=1/(-5/3)
cos Ф=-3/5
Therefore:
tan(Ф/2)=-√[(1-cosФ)/(1+cosФ)]
tan(Ф/2)=-√[(1+3/5)/(1-3/5)]
tan(Ф/2)=-√[(8/5)/(2/5)]
tan(Ф/2)=-√4
tan(Ф/2)=-2
Answer: tan (Ф/2)=-2; when tan (Ф)=4/3
Relative extrema occur where the derivative is zero (at least for your polynomial function).
So taking the derivative we get
<span>20<span>x3</span>−3<span>x2</span>+6=0
</span><span>
This is a 3rd degree equation, now if we are working with complex numbers this equation is guaranteed to have 3 solutions by the fundamental theorem of algebra. But the number of real roots are 1 which can be found out by using Descartes' rule of signs. So the maximum number of relative extrema are 1.</span>
Answer:
6983/1000 OR 6.983
Step-by-step explanation:
6*1=6
9*1/10=9/10
8*1/100=8/100
3*1/1000=3/1000
6+9/10+8/100+3/1000=6983/1000
OR
6.983
Answer:
i beleive it is the 3rd choice
Step-by-step explanation:
