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Trigonometric Identiity

Date: 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   
Associated Topics:
High School Trigonometry

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