Proof of a Trig IdentityDate: 03/13/98 at 20:03:02 From: Chris Hebert Subject: Proving Trigonomic Identies How can I prove that 1 + sinX + cosX 1 + cosX ------------------- = ----------- 1 + sinX - cosX sinX I have tried solving this problem using Basic Trigonometric Identities, Pythagorean Identities, Double-Angle Formulas, and Half-Angle Formulas. This problem has stumped the entire Pre-Calculus class, including my teacher! Any help would be appreciated. Even if this problem stumps you, will you please E-mail and tell me so, or let me know any progress you have made? (I do know that this problem is solvable though, because it was published on a work sheet). Date: 03/16/98 at 17:14:13 From: Doctor Fred Subject: Re: Proving Trigonomic Identies Proof: Often, multiplying by conjugates is helpful in trig identities. 1 + sinX + cosX (1 + sinX + cosX)^2 ---------------- = ----------------------------------- 1 + sinX - cosX (1 + sinX - cosX)(1 + sinX + cosX) 1 + 2sinX + 2cosX + 2sinXcosX + (sinX)^2 + (cosX)^2 = ----------------------------------------------------- 1 + 2sinX + (sinX)^2 - (cosX)^2 Now, notice almost everything has a coefficient of 2. Substituting 1 for (sinX)^2 + (cosX)^2 in the numerator and (sinX)^2 for 1 - (cosX)^2 in the denominator gives 2 + 2sinX + 2cosX + 2sinXcosX = ------------------------------- 2sinX + 2(sinX)^2 Factor by grouping in the numerator, and factor out sinX in the denominator (1 + sinX)(1 + cosX) = ---------------------- (1 + sinX)sinX Dividing numerator and denominator by 1+sinX gives 1 + cosX = ----------- sinX -Doctor Fred, The Math Forum http://mathforum.org/dr.math/ |
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