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Using implicit differentiation, let y equal arccos(x) : y=arccos(x). So x = cos(y), and dx/dy = -sin(y). dy/dx is therefore -1/sin(y). from trig indentities: sin(y) = sqrt(1-cos^2(y)). Substituting gives ...

Since x=5+i is a solution to f(x)=0 we then know that x=5-i must also be a solution to f(x)=0, by the complex conjugate root theorem.Now we can break f down into the product of a polynomial and these two ...

For a polynomial with real coefficients, use that roots come in complex conjugate pairs. Therefore, another root is 2-i (and we know for this example that the final root must be real). Write the factorise...

First of all, replace sinxcosx with 1/2 sin2x. Then you should let U=1/2 Sin2x and replace that in the formula. If y=arctan(U), then U=tany. work out dU/dy which is Sec²y. Using the trigonometr...

First, recall how to construct a proof by simple induction in this manner: (1) Assume statement true for n=k, (2) Prove true for n=k+1, (3) Show true for n=1.(1) => ∑(from r=1 to k) r(2r−1)(3r−1)= (k/6...