S Stephxox New member Joined Sep 28, 2008 Messages 6 Nov 12, 2008 #1 question: 1/cosx -cosx = sinx tanx This is what I have so far: RS=sinx (sinx/cosx) =sin2x/cosx =1-cos2x/cosx =1-cosx Where do I go from here? I'm Stuck
question: 1/cosx -cosx = sinx tanx This is what I have so far: RS=sinx (sinx/cosx) =sin2x/cosx =1-cos2x/cosx =1-cosx Where do I go from here? I'm Stuck
P peekaboos New member Joined Nov 12, 2008 Messages 1 Nov 12, 2008 #2 Re: How Do you prove this trig identity? You're stuck because this is where you went wrong. (in bold) RS=sinx (sinx/cosx) =sin2x/cosx This is wrong =1-cos2x/cosx =1-cosx sinx(sinx/cosx) is equaled to sin^2x/cosx unless you forgot the exponent - not sure. Anyway the answer is: LS 1/cosx - cosx Find common denominator =1-cosx/cosx Then 1-cosx is sin^2 by using pythagorean identity. =sin^2x/cosx RS sinx(sinx/cosx) =sin^2/cosx LS = RS, identity proven.
Re: How Do you prove this trig identity? You're stuck because this is where you went wrong. (in bold) RS=sinx (sinx/cosx) =sin2x/cosx This is wrong =1-cos2x/cosx =1-cosx sinx(sinx/cosx) is equaled to sin^2x/cosx unless you forgot the exponent - not sure. Anyway the answer is: LS 1/cosx - cosx Find common denominator =1-cosx/cosx Then 1-cosx is sin^2 by using pythagorean identity. =sin^2x/cosx RS sinx(sinx/cosx) =sin^2/cosx LS = RS, identity proven.
