Question

find the principle value ​

Answer 1

$\boxed{\textbf{\large{To Find}}}$

principal values of

1) sin^(-1 )( 1/ 2 )+ cos^(-1 ) (1 / 2)

2) sin^(-1 )(sin (3π / 5 ))

$\boxed{\textbf{\large{Explanation}}}$

Q1) sin^(-1 )( 1/ 2 )+ cos^(-1 ) (1/2)

for, sin^(-1)(1/2)

let , sin^(-1)(1/2 ) = y

sin y = 1/2 =cos π/6

principal value branch for sin^(-1) is [ -π/2 , π / 2] and -π/2 < π /6 < π/ 2

so, y = π / 6

sin^(-1 ) ( 1/ 2) = π / 6 ..............(1)

Now,

for, cos^( -1)(1/ 2)

let, cos^(-1)(1/2) = x

cosx = 1/2 = cos π/3

principal value branch for cos^(-1) is [ 0 , π ] and 0 < π /3 < π

x = π / 3

cos^(- 1) ( 1/2 ) = π /3 ...............(2)

from (1) and (2)

sin ^( -1 )( 1/ 3) + cos^( -1)( 1 / 2)

= (π / 6) + ( π / 3 )

=(( 3π + 6π ) / 18 )

= ( 9π / 18 )

= π / 2

principal value of sin^(-1 )( 1/ 2 )+ cos^(-1 ) (1 / 2) is

$\boxed{\textbf{\large{pie / 2}}}$

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Q2) sin^(-1 )(sin (3π / 5 ))

As we know the range of principle value branch of sin^( -1 ) is [-π /2 , π / 2]

let us suppose,

sin^( -1 ) ( sin( 3π /5) )

= sin^( -1 ) [ sin ((5π - 2π) / 5 )]

= sin^( -1 ) [ sin ( π -( 2π / 5))]

Therefor,

sin^( -1) ( sin 3π /5 )

= sin^(-1 ) (sin (2π /5))

we know, ( 2π / 5 )€ [ -π / 2 , π /2 ]

sin^( -1 )( sin (3π/5 )) = 2π / 5

Therefor the principle value of

sin^(-1 )(sin (3π / 5 )) is

$\boxed{\textbf{\large{2pie / 5}}}$

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Answer 2

welcome to the concept of inverse trigonometry......

⏩ I hope it help you ❤️