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
Answer:
??
Step-by-step explanation:
The experiment with the least number of trials.
Experimental probability is more accurate and more close to theoretical probability by having the most trials. More trials = more accuracy. Less trials = less accuracy.
Answer:
-40
Explanation:
Given
u = 2i - j; v= -5i + 4j and w = j
Required
4u(v-w)
4u = 4(2i) = 8i
v - w = -5i + 4j - j
v - w = -5i + 3j
Substitute
4u(v-w)
= 8i(-5i+3j)
= -40(i*i) [since i*i = 1]
= -40
Hence the required solution is -40
Y = kx where k is a constant
27 = k*9
k = 27/9 = 3
required variation is y = 3x
