Transcript Values Involving Inverse Trigonometric Functions II

Values Involving Inverse
Trigonometric Functions II
Finding arctrg(trgθ)
Examples I
Arcsin
arcsin[sin( - π/ 4 )] = - π/ 4
Notice that:
- π/ 4 belongs to the range of the function
y = arcsinx,
which is [ -π/2 , π/2 ]
Arctan
arctan[tan( - π/ 4 )] = - π/ 4
Notice that:
- π/ 4 belongs to the range of the function
y = arctanx,
which is ( -π/2 , π/2 )
Arccos
arccos[cos( 3π/ 4 )] = 3π/ 4
Notice that:
3π/ 4 belongs to the range of the function
y = arccosx,
which is [ 0 , π ]
Arccot
arccot[cot( 3π/ 4 )] = 3π/ 4
Notice that:
3π/ 4 belongs to the range of the function
y = arccotx,
which is ( 0 , π )
Arcsec
arcsec[sec( 5π/ 4 )] = 5π/ 4
Notice that:
5π/ 4 belongs to the range of the function
y = arcsecx,
which is [ 0 , π/2 ) U [π , 3π/2 )
Examples II
Arcsin
arcsin[sin( 3π/ 4 )]
Notice that:
3π/ 4 does not belong to the range of the
function y = arcsinx, which is [ -π/2 , π/2 ]
sin( 3π/ 4 ) = 1/√2 ( Why? )
Thus,
arcsin[sin( 3π/ 4 )] = arcsin[1/√2] = π/ 4
Arctan
arctan[tan( 7π/ 4 )]
Notice that:
7π/ 4 does not belong to the range of the
function y = arctanx, which is ( -π/2 , π/2 )
tan( 7π/ 4 ) = -1 ( Why? )
Thus,
arctan[tan( 7π/ 4 )] = arcsin[-1] = -π/ 4
Arccos
arccos[cos( - π/ 4 )]
Notice that:
-π/ 4 does not belong to the range of the
function y = arccosx, which is [ 0 , π ]
cos( - π/ 4 ) = 1/√2 ( Why? )
Thus,
arccos[cos( -π/ 4 )] = arccos[1/√2] = π/ 4
Arccot
arccot[cot( 7π/ 4 )]
Notice that:
7π/ 4 does not belong to the range of the
function y = arccotx, which is ( 0 , π )
cot( 7π/ 4 ) = -1 ( Why? )
Thus,
arccot[cot( 7π/ 4 )] = arccot[-1]
= 3π/ 4
Arcsec
arcsec[sec( 3π/ 4 )]
Notice that:
3π/ 4 does not belong to the range of the
function y = arcsecx,
which is [ 0 , π/2 )U[π,3π/2)
sec( 3π/ 4 ) = - √2 ( Why? )
Thus,
arcsec[sec( 3π/ 4 )] = arcsec[ - √2]
= 5π/ 4