Search over 10,000 free study notes

We can use the characteristic equation, det(A - kI) = 0 to find the eigenvalues of A. Performing this, we see that(2-k) * ( (1-k)(3-k) - 3 ) = 0.Immediately, we can see a root is k = 2, which is ...

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...