For any angle - 0-sin- -1- then sin2-

For any angle θ∈ (0, π], sinθ =1 Now, we know sinθ =1 for θ=π2,5π2,9π2, ... nbsp; And, since θ∈(0,π]⇒θ=π2 ⇒sin2θ=sin(2×π2)=sin(π) Now, we know at θ =π, sinθ = ...

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Standard XII

Mathematics

Relation between Inverses of Trigonometric Functions and Their Reciprocal Functions

For any angle...

Question

For any angle $θ \in (0, π], sin θ = 1,$ then $sin 2 θ =$

0

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\frac{1}{\sqrt{2}}

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Solution

The correct option is A $0$
For any angle $θ \in (0, π], sin θ = 1$
Now, we know $sin θ = 1 for θ = \frac{π}{2}, \frac{5 π}{2}, \frac{9 π}{2}, . . .$
And, since $θ \in (0, π] \Rightarrow θ = \frac{π}{2}$
$\Rightarrow sin 2 θ = sin (2 \times \frac{π}{2}) = sin (π)$
Now, we know at $θ = π, sin θ = 0$

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For any angle θ∈(0,π],sinθ=1, then sin2θ=

The correct option is A 0For any angle θ∈(0,π],sinθ=1 Now, we know sinθ=1 for θ=π2,5π2,9π2,... And, since θ∈(0,π]⇒θ=π2 ⇒sin2θ=sin(2×π2)=sin(π) Now, we know at θ=π,sinθ=0

For any angle $θ \in (0, π], sin θ = 1,$ then $sin 2 θ =$