PRECAL!
posted by just .
Prove: (3cosx + 4sinx)^2 + (4cosx  3sinx)^2=5 ...this is all under the square root sign.
this is what i did please check.
i multipied both binomials. but i get cos square x..and sincosx ..how do i combine them..please help!
In this case, after multiplying binomials, there will be 24sincos and 24sincos terms that cancel. That will leave you with sin^2 and cos^2 terms. Write sin^2 as 1cos^2 and solve for cosx
i get 9cos^2x + 16sin^2x +16cos^2x + 9sin^2x
now i know that sin^2 equals 1cos^2....so would that be 9cos^2x + 17cos^2x+16cos^2+ 10cos^2x? if not please help..and if this is correct..what would come after this step?
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Prove: (3cosx + 4sinx)^2 + (4cosx  3sinx)^2=5 ...this is all under the square root sign. I get 9cos^2x + 16sin^2x +16cos^2x + 9sin^2x now i know that sin^2 equals 1cos^2....so would that be 9cos^2x + 17cos^2x+16cos^2+ 10cos^2x? 
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