find sinx/2, cosx/2 and tanx/2. 1. cosx= - 1/3, x in quadrant 3
Answers
Answer:
my teachers answer not mine!\
Step-by-step explanation:
1)Cos(x)=−13, x is in quad 3
Using double angle formula :
Cos(x)=2cos2(x2)−1=1−sin2(x2)
Cos(x)=−13=2cos2(x2)−1
2cos2(x2)=−13+1
cos2(x2)=−13+12
cos2(x2)=13
cos(x2)=±1√3
Cos(x)=−13=1−2sin2(x2)
2sin2(x2)=1−−13
sin2(x2)=1+132
sin2(x2)=46
sin(x2)=±√(23)
Tan(x2)=sin(x2)cos(x2)=±√(23)±1√3
Tan(x2)=±√2
Given that x lies in quad 3. i.e, it lies between 180°and 270°. 180°>x>270°. Then, for x2:180°2>x2>270°2=90°>x2>135°. Hence x2 lies in quad 2. In quad 2 only sine has positive value.
Therefore we have following:
cos(x2)=−1√3
sin(x2)=√(23)
Tan(x2)=−√2
….
2)tan(x)=34,x is in the quad 3.
Here, again we use double angle formula:
Tan(x)=2tan(x2)1−tan2(x2)
Tan(x)=34=2tan(x2)1−tan2(x2)
34(1−tan2(x2))−2tan(x2)=0
34−34(tan2(x2))−2tan(x2)=0
Let tan(x2)=t.
Then,34−34(tan2(x2))−2tan(x2)=34−34(t2)−2t=0
Thus we have a quadratic equation whose solution is given by
t=−b±√(b2−4ac)2a=−b±√D2a
where, discriminant, D=b2−4ac decides the number and nature if the solution of t.
D=b2−4ac=82−4∗(3)(−3)=64−(−36)=100
Since Dis positive the solution will be two distinct real roots.
tan(x2)=t=−b±√D2a=−8±√1002(3)
tan(x2)=−8±106=13,−3
As we have already seen above that x is quad 3then x2is in quad 2 where tan(x) is negative. Thus,tan(x2)=−3
cos(x)=1−tan2(x2)1+tan2(x2)=1−(−3)21+(−3)2
Cos(x)=1−91+9=−45
Then, as in the above we can obtain cos(x2) and sin(x2)
Cos(x)=−45=2cos2(x2)−1
2cos2(x2)=−45+1
cos2(x2)=−45+12
cos2(x2)=110
cos(x2)=±1√10
Since, x2is in quad 2 we know that cosine is also negative. Thus,cos(x2)=−1√10
Cos(x)=−45=1−2sin2(x2)
2sin2(x2)=1−−45
sin2(x2)=1+452
sin2(x2)=910
sin(x2)=±(3√10)
Since,x2is in quad 2 we know that sine is positive. Thus,sin(x2)=(3√10).
Therefore we have following:
tan(x2)=−3
cos(x2)=−1√10
sin(x2)=(3√10)