MaxWong

\(\log_86\approx0.861654166907\)

\(\frac{1}{1-x-x^2}\\\)Step 1

\(= \frac{1}{(x-\frac{1+\sqrt5}{2})(x-\frac{1-\sqrt5}{2})}\)Step 2

\(=\frac{x+\frac{1+\sqrt5}{2}}{(x-\frac{1+\sqrt5}{2})(x+\frac{1+\sqrt5}{2})}\times \frac{x+\frac{1-\sqrt{5}}{2}}{(x-\frac{1-\sqrt5}{2})(x+\frac{1-\sqrt{5}}{2})}\)Step 3

\(=\frac{2x+1+\sqrt5}{2x^2-(3+\sqrt{5})}\times \frac{2x+1-\sqrt{5}}{2x^2-(3-\sqrt{5})}\)Step 4

\(=\frac{((2x+1)+\sqrt5)((2x+1)-\sqrt5)}{((2x^2-3)-\sqrt{5})(((2x^2-3)+\sqrt{5}))}\)Step 5

\(=\frac{(2x+1)^2-{\sqrt5}^2}{(2x^2-3)^2-{\sqrt5}^2}\)Step 6

\(=\frac{4x^2+4x+1-5}{4x^4-12x^2+9-5}\)Step 7

\(=\frac{4x^2+4x-4}{4x^4-12x^2+4}\)Step 8

\(=\frac{x^2+x-4}{x^4-3x^2+1}\)

The coefficient is generated because the denominator of the 2 fractions in step 3 involves a fraction with denominator 2. So that 2 is multiplied in step 4 so the coefficient is generated.

\(\tan t=\frac{12}{5}\\ hypotenuse = \sqrt{5^2+12^2}=13\\ \cos \space t= \frac{5}{13} \\\quad\sin(\frac{t}{2})\\= \pm\sqrt{\frac{1-\cos t}{2}}\\=\pm\sqrt{\frac{1-\frac{5}{13}}{2}}\\=\pm\sqrt{\frac{\frac{8}{2}}{13}}\\\)

\(=\pm\sqrt{\frac{4}{13}}\\=\frac{\pm2}{\sqrt{13}}\)

\(\because \space \tan t \space is \space positive\\\therefore \sin(\frac{t}{2})\space is\space positive\space too\)

\(\therefore \sin(\frac{t}{2})=\frac{2\sqrt{13}}{13}\space OR \frac{2}{\sqrt{13}}\)

By the way, why pie?

(a^2 + b^2 + c^2)(b^2 + c^2 + d^2) always equals (ab + bc + cd)^2

In GP or not in GP does not matter.

3. (ab + bc + cd)^2

1.98892\(\times\)1030=2048.5876

I think you mean cos^4 x + sin^2 x . cos^2 x

\(cos^4x+sin^2xcos^2x\\=cos^2x(cos^2x+sin^2x)=cos^2x\)

\(\quad5(3n^2-n+4)^3\\=135n^6-135n^5+585n^4-365n^3+780n^2-240n+320\)

Ya we do need more Russian users.

7/14 = 1/2

According to special angles table, cos \(60^{\circ}\) = \(\frac{1}{2}\)

\(\therefore cos^{-1}(\frac{7}{14})=60^{\circ}\)

5 mod 3 = (the remainder of \(5\div 3\))= 2