Math, asked by bsridevi1981, 9 months ago

theta belongs to (pi,3pi/2) then root(1-sin^2theta) is

Answers

Answered by Lueenu22
0

Step-by-step explanation:

The sum of all values of \thetaθ is \boxed{2\pi}

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

+(−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

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

Putting n = -2

\theta=-\frac{2\pi}{3}+\frac{\pi}{9}=-\frac{5\pi}{9}θ=−

3

+

9

π

=−

9

which does not belong in [0, \pi][0,π]

Putting n = 2

\theta=\frac{2\pi}{3}+\frac{\pi}{9}θ=

3

+

9

π

\implies \theta=\frac{7\pi}{9}⟹θ=

9

Putting n = 3

\theta=\frac{3\pi}{3}-\frac{\pi}{9}θ=

3

9

π

\implies \theta=\frac{8\pi}{9}⟹θ=

9

Putting n = 4

\theta=\frac{4\pi}{3}+\frac{\pi}{9}θ=

3

+

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

,

9

,

9

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

+

9

+

9

S=\frac{18\pi}{9}S=

9

18π

\implies S=2\pi⟹S=2π

Hope this answer is helpful.

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