Values Involving Inverse Trigonometric Functions II Finding arctrg(trgθ)

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Presentation transcript:

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