Question

arccos(0) = π/2 because cos(π/2) = 0 and π/2 is within range of arccos which is [0 , π] arcsin(-1) = -π/2 because sin(-π/2)​

Answer 1

Answer:

sin(arccos(−

2

1

))=sin[cos

−1

(−

2

1

)]

Let cos

−1

(−

2

1

)=P

cosp=−

2

1

⇒P=π−

3

π

=

3

2π

=sin(

3

2π

)

=sinπ−

3

π

=sin

3

π

sin(arccos(−

2

1

))=

2

3

∴sin(arccos(−

2

1

))=

2

3

Answer 2

Answer:

The cosine function has a range of [−1,1] , meaning the inverse cosine function has a domain of [−1,1] . The reason cos(arccos(3π)) is undefined is because arccos(3π) is undefined, as 3π is outside of the domain for arccos . ... arccos(cos(2π))=arccos(1) , which is a value such that cos(arccos(1))=1