Question

The original question is to write 2x^2 - 4xy - y^2 + 8 = 0 in canonical form.
I am struggling to solve cos((1/2)arctan(4/3))

Answer 1

Just use a calculator...........

Answer 2

@oldrin.bataku I don't understand the question, please explain me

Answer 3

my question is they are separate problem or just one? what is the link?

Answer 4

I need help with the first part of the question

Answer 5

$$\cos\left(\frac12\arctan(4/3)\right)=\cos\frac12\theta=\sqrt{\frac{1+\cos\theta}2}$$Clearly \(\cos\theta=3/5\) so $$\cos\frac12\theta=\sqrt{\frac{1+3/5}2}=\sqrt{4/5}=\frac2{\sqrt5}$$