Question

Sin^-1(sin47pie/7)=? And the sun inverse does not get cancelled with sine pls explain the property

Answer 1

Given ---> Sin⁻¹ { Sin ( 47π / 7 ) }

Your question is why sin⁻¹ not cancelled with Sin

We know that domain and range of sin⁻¹ are

Domain = [ -1 , 1 ] , Range = [-π/2 , π/2 ]

( 47π / 7 ) is not in the range of sin⁻¹ so we can not discard Sin⁻¹ by Sin in this case

Now to solve this problem, first we solve Sin( 47π / 7 )

Sin( 47π / 7 ) = Sin { 6π + (5π /7) }

We know that Sin( 2nπ + θ ) = Sinθ , applying it we get

= Sin ( 5π / 7 )

We know that Sin ( π - θ ) = Sinθ , applying it we get

= Sin { π - ( 2π / 7 ) }

= Sin ( 2π / 7 )

Now

Sin⁻¹ {Sin(47π /7)} = Sin⁻¹{ Sin( 2π / 7)}

= ( 2π / 7 )

Additional information---->

(1) Doain of Cos⁻¹ , [-1 , 1 ]

Range of Cos⁻¹ [0 , π ]

(2) Domain of tan⁻¹ , R

Range of tan⁻¹ , ( -π/2 , π / 2 )

(3) Domain of Cosec⁻¹ (-∞ , -1]U[1 , ∞ )

Range of Cosec⁻¹ [-π/2 , π/2 ]- {0}

(4) Domain of Sec⁻¹ (-∞,-1]U[1,∞)

Range of Sec⁻¹ [0, π] - { π/2 }

(5) Domain of Cot⁻¹ R

Range of Cot⁻¹ ( 0 , π )

Answer 2

Step-by-step explanation:

Given ---> Sin⁻¹ { Sin ( 47π / 7 ) }

Your question is why sin⁻¹ not cancelled with Sin

We know that domain and range of sin⁻¹ are

Domain = [ -1 , 1 ] , Range = [-π/2 , π/2 ]

( 47π / 7 ) is not in the range of sin⁻¹ so we can not discard Sin⁻¹ by Sin in this case

Now to solve this problem, first we solve Sin( 47π / 7 )

Sin( 47π / 7 ) = Sin { 6π + (5π /7) }

We know that Sin( 2nπ + θ ) = Sinθ , applying it we get

= Sin ( 5π / 7 )

We know that Sin ( π - θ ) = Sinθ , applying it we get

= Sin { π - ( 2π / 7 ) }

= Sin ( 2π / 7 )

Now

Sin⁻¹ {Sin(47π /7)} = Sin⁻¹{ Sin( 2π / 7)}

= ( 2π / 7 )