Hi there, I need help proving a trig ident. steps plz, cos2x= (1-tan^2(x)) ..........(1+tan^2(x)) tan^2(x) is tan(x)^2....if u don't understand my writing Please, any help is appreciated, thanks
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You can use the identity: $\displaystyle tan^{2}(x)=\frac{1-cos(2x)}{1+cos(2x)}$ It'll fall into place.
thanks so much
Originally Posted by compuwiz cos2x= (1-tan^2(x)) ..........(1+tan^2(x)) $\displaystyle 1+\tan^2x=\sec^2x.$ (Basic identity.) Now split the original fraction into two sums: $\displaystyle \frac{{1 - \tan ^2 x}} {{\sec ^2 x}} = \cos ^2 x - \sin ^2 x.$ And this is exactly equal to double-angle cosine formula.
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