Sin2x + _sin2x = 2 1-cosx 1+cosx i used sin^2x + cos^2 x = 1 then got (sin^2x)/(1-cos^2x)=0 and thats a diff of 2 squares so its (sin^2x)/(1-cosx)(1+cosx) = 0 but where does the two come from what did i do wrong??
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Hello deej813 Originally Posted by deej813 Sin2x + _sin2x = 2 1-cosx 1+cosx i used sin^2x + cos^2 x = 1 then got (sin^2x)/(1-cos^2x)=0 and thats a diff of 2 squares so its (sin^2x)/(1-cosx)(1+cosx) = 0 but where does the two come from what did i do wrong?? I'm not clear what the question is here. My best guess is that you have to solve the equation If that's it then you'd write: , using Grandad
nah i've gotta prove that the LHS = RHS = 2 so rearrange the LHS and get it to equal two but i keep getting 0
Hello deej813 Sorry, but the LHS just isn't identically equal to 2. If you followed my working you'd see that it's , and that's as far as you can go. So here it is once again: Grandad
sorry its sin^2x on top of both fractions. does that make a difference??
Hello deej813 Originally Posted by deej813 sorry its sin^2x on top of both fractions. does that make a difference?? It sure does! Instead of on the top then, we get: Grandad
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