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Monday, April 27, 2009

exact vaue of sin(arctan(4/3)-arccos(12/13))

exact value of sin [arctan(4/3)-arccos(12/13)]

let A = arctan(4/3)
tanA = 4/3
draw a rough right triangle to get

sinA = 4/5 ; cosA = 3/5



let B = arccos(12/13)
as before cosB = 12/13 ; sinB = 5/13

now sin [arctan(4/3)-arccos(12/13)] = sin[A-B] = sinAcosB - cosAsinB trigonometric identities
=(4/5 )(12/13) - (3/5)(5/13) = 33 / 65


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