theta belongs to (pi,3pi/2) then root(1-sin^2theta) is
Answers
Step-by-step explanation:
The sum of all values of \thetaθ is \boxed{2\pi}
2π
Step-by-step explanation:
\sin 3\theta=\frac{\sqrt{3}}{2}sin3θ=
2
3
\implies \sin 3\theta=\sin\frac{\pi}{3}⟹sin3θ=sin
3
π
..... (1)
We know that if \sin\theta=\sin\alphasinθ=sinα
Then the general value of \thetaθ is given by
\theta=n\pi+(-1)^n\alphaθ=nπ+(−1)
n
α
Where n=0,\pm 1, \pm 2, \pm 3, ......n=0,±1,±2,±3,......
Thus, the general value of (1) is
3\theta=n\pi+(-1)^n\frac{\pi}{3}3θ=nπ+(−1)
n
3
π
\implies \theta=\frac{n\pi}{3}+(-1)^n\frac{\pi}{9}⟹θ=
3
nπ
+(−1)
n
9
π
Putting n = 0
\theta=\frac{\pi}{9}θ=
9
π
Putting n = -1
\theta=-\frac{\pi}{3}-\frac{\pi}{9}=-\frac{4\pi}{9}θ=−
3
π
−
9
π
=−
9
4π
which does not belong in [0, \pi][0,π]
Putting n = 1
\theta=\frac{\pi}{3}-\frac{\pi}{9}θ=
3
π
−
9
π
\implies \theta=\frac{2\pi}{9}⟹θ=
9
2π
Putting n = -2
\theta=-\frac{2\pi}{3}+\frac{\pi}{9}=-\frac{5\pi}{9}θ=−
3
2π
+
9
π
=−
9
5π
which does not belong in [0, \pi][0,π]
Putting n = 2
\theta=\frac{2\pi}{3}+\frac{\pi}{9}θ=
3
2π
+
9
π
\implies \theta=\frac{7\pi}{9}⟹θ=
9
7π
Putting n = 3
\theta=\frac{3\pi}{3}-\frac{\pi}{9}θ=
3
3π
−
9
π
\implies \theta=\frac{8\pi}{9}⟹θ=
9
8π
Putting n = 4
\theta=\frac{4\pi}{3}+\frac{\pi}{9}θ=
3
4π
+
9
π
\implies \theta=\frac{13\pi}{9}⟹θ=
9
13π
which does not belong in [0, \pi][0,π]
Therefore, the velues of \thetaθ belonging in [0, \pi][0,π] are
\theta=\frac{\pi}{9}, \frac{2\pi}{9}, \frac{7\pi}{9}, \frac{8\pi}{9}θ=
9
π
,
9
2π
,
9
7π
,
9
8π
The sum of the all the values is
S=\frac{\pi}{9}+\frac{2\pi}{9}+\frac{7\pi}{9}+\frac{8\pi}{9}S=
9
π
+
9
2π
+
9
7π
+
9
8π
S=\frac{18\pi}{9}S=
9
18π
\implies S=2\pi⟹S=2π
Hope this answer is helpful.
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