Melody

$$\\2cos^2\theta=1\\\\
cos^2\theta=1/2\\\\
cos\theta=\pm \;1/\sqrt{2}\\\\
\theta = 45^0+\frac{n\pi}{4}\qquad N\in Z\qquad $that means N is any integer$\\\\$$

It is a function with a cubic in it.

for instance

$$\\y=x^3\\
y=x^3-7\\
y=(x+5)^3$$

etc

You need brackets Eloise :)

Anon, what do you want us to do with this?

Does   $$j=\sqrt{-1}$$      or is it just an ordinary pronumeral?

Thanks Chris, if I get up to it I would like to take a better look at this question.  

It looks like it would take some effort to get ones head around it.

I hope anon says thank you to you :)

YES that looks really good MG.

There are a couple of points I'd like to make though.  

First, you did not simplify  $$\frac{4}{10}$$    right at the very begining!

second, you are still splitting it into bits - i don't wnat you to do that!  

I accept that not ALL your working will be shown, some will be scribbled on bits of paper and it does not need to be included.  

Also, when you are adding and subtracting mixed numerals do not change them into improper fractions.

It is NOT necessary and it makes the calculations worse!

$$\\1\frac{3}{4}\times 7 +8^2-6\div \frac{4}{10}\\\\
1\frac{3}{4}\times 7 +8^2-6\div \textcolor[rgb]{1,0,0}{\frac{2}{5}}\\\\
=1\frac{3}{4}\times7+\textcolor[rgb]{1,0,0}{64} -6\div\frac{2}{5}\qquad\\\\
=\textcolor[rgb]{1,0,0}{\frac{4*1+3}{4}}\times7+64 -6\div\frac{2}{5}\qquad\\\\
=\textcolor[rgb]{1,0,0}{\frac{7}{4}}\times\textcolor[rgb]{1,0,0}{\frac{7}{1}}}+64 -6\div\frac{2}{5}\qquad\\\\
=\textcolor[rgb]{1,0,0}{\frac{49}{4}}+64 -6\div\frac{2}{5}\\\\
=\frac{49}{4} +64-6\textcolor[rgb]{1,0,0}{\times\frac{5}{2}}\\\\
=\frac{49}{4} +64-\textcolor[rgb]{1,0,0}{\frac{6}{1}\times\frac{5}{2}}\\\\
=\frac{49}{4} +64-\textcolor[rgb]{1,0,0}{\frac{3}{1}\times\frac{5}{1}}\\\\
=\frac{49}{4} +64-\textcolor[rgb]{1,0,0}{\frac{3*5}{1*1}}\\\\$$

$$\\=\frac{49}{4} +64-\textcolor[rgb]{1,0,0}{15}}\\\\
=\textcolor[rgb]{1,0,0}{12\frac{1}{4}} +64-15\\\\
=\textcolor[rgb]{1,0,0}{12+\frac{1}{4}} +64-15\\\\
=\textcolor[rgb]{1,0,0}{64+12-15+\frac{1}{4}}\\\\
=\textcolor[rgb]{1,0,0}{64-3+\frac{1}{4}}\\\\
=\textcolor[rgb]{1,0,0}{61+\frac{1}{4}}\\\\
=61\frac{1}{4}$$

Now that was an effort!

Look though how I have done it MG and learn!!   I am sure that you will :)

Especially look at the mixed numeral addition!

I am rusty but I will give it a go.

find two possible values of p if the lines px-y+1=0 and 3x+y+1=0 intersect at 45 degrees

px-y+1=0             y=px+1             gradient =p

this line makes an angle of    atan(p)    with the positive axis    I'll call this  $$\theta_1$$

3x+y+1=0            y=-3x-1            gradient = -3

this line makes an angle of    atan(-3)    with the positive axis     I'll call this  $$\theta_2$$

Now we want

 $$\theta_1-\theta_2 = \pm \frac{\pi}{4}$$

You need to use the last formula, do you think you can try to take it from here :)

ROF LOL         

Reposted here

This is a repost

original question is here

$$\\Sin^2(280+2\pi) \times Log(Cos85) + tan(\pi/2) - \sqrt{567}\\
Sin^2(280) \times Log(Cos85) + 1 - \sqrt{567}\\$$

$${\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{280}}^\circ\right)}}^{\,{\mathtt{2}}}{\mathtt{\,\times\,}}{log}_{10}\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{85}}^\circ\right)}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{1}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{567}}}} = -{\mathtt{23.839\: \!511\: \!806\: \!008\: \!947\: \!7}}$$