Trigonometric IdentiityDate: 07/29/99 at 22:48:48 From: Stephen Sisk Subject: Proof of a Trigonometric Identiity While proving arcsin(-x)= - arcsin(x), my initial thought was to let x = sin(x) and use similar identity for sines. Is this a valid argument? If not, how would you show this? Date: 07/30/99 at 10:25:39 From: Doctor Rick Subject: Re: Proof of a Trigonometric Identiity You are on the right track, though you will confuse yourself if you try to make the same variable mean two things (a number and its sine). Try this: y = arcsin(x) <=> x = sin(y) x = sin(y) = -sin(-y) You can complete the work to show that y = -arcsin(-x) However, there is a complication. The inverses of the trigonometric functions are not functions because they are multiple-valued. To make them into the functions arcsin, arccos, etc., we restrict them to the principal value: in the case of arcsin, this is -pi/2 <= arcsin(x) <= pi/2 In other words, the first line of my proof outline above should be y = arcsin(x) <=> x = sin(y) AND -pi/2 <= y <= pi/2 You need to include the range similarly when you switch back from sine to arc sine. It works out easily enough, but it is just something you cannot ignore. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
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