Melody

Find sin θ if cot θ = - 2 and cos θ < 0.

Thanks Omi for that great answer, 

I have yet another method.

Since cos is negative the answer must be in the 2nd or third quad

Since cot is neg, tan is also neg so it is in the 2nd or  4th quad.

Therefore it is in the 2nd quad where sine is positive!

I am going to put that knowledge aside for a bit and work out the first quadrant equivalent.

\(cot\theta=\frac{2}{1}\\ so\\ tan\theta=\frac{1}{2}\)

Draw a right angled triangle and mark one angle as theta, then put

1 on the opposite side and

2 on the adjacent side so that tan theta = 1/2

Find the hypotenuse using pythagoras   

\(hypotenuse=\sqrt{1^2+2^2}=\sqrt{5}\)

\(sin\theta = \frac{opp}{hyp}=\frac{1}{\sqrt5}=+\frac{\sqrt5}{5}\)

(positive for our final answer too because theta is in  the 2nd quad where sine is positive.)

At this point is is FINISHED.  You are not asked to find theta :))

so

These two attitudes should not be bundles together!

OK you find maths is hard and you most likely dislike it.

That is ok, there are lots of things I dislike too, just maths is not one of them.

If life suxs for you that has nothing to do with maths.  If you think it does then it is your thoughts that are the problem, not the mathematics.

------------

A suggestion for you.  If you want to do better at maths and do not want it to totally stress you out then put time into practicing and fully understanding the easier stuff.  Do not get too stressed about things that are too difficult.  Maybe you will even start not disliking it too much :)

For people who excel at school and want to have highly professional careers, school results are very important.  For other people a few marks more or less makes little difference.  Do the best you reasonably can but don't let it make you miserable!  There is MUCH more to life than school.

\((a^{-1}+b^{-1})(a^{-1}-b^{-1})^3\\ =\left(\frac{1}{a}+\frac{1}{b}\right)\times\left(\frac{1}{a}-\frac{1}{b}\right)^3\\ =\left(\frac{b}{ab}+\frac{a}{ab}\right)\times\left(\frac{b}{ab}-\frac{a}{ab}\right)^3\\ =\left(\frac{b+a}{ab}\right)\times\left(\frac{b-a}{ab}\right)^3\\ =\frac{(b+a)(b-a)^3}{a^4b^4}\\ \)

You may finish here but you may want to expand it.  

\(=\frac{(b+a)(b-a)\times(b-a)^2}{a^4b^4}\\ =\frac{(b^2-a^2)\times(b^2-2ab+a^2)}{a^4b^4}\\ =\frac{b^4-2ab^3+2a^3b-a^4}{a^4b^4}\\ \)

You can factorise this but you cannot simplify it.

Well

70% of 80% of original is $28

You need to turn this into an equation and then solve it :)

Hint

Let original be X

There is a calc on the home page.  

Off you go, give your fingers some more exercise :)

What do you want to divide it by?

Can't you just put it into the calculator?

If you needto use a formula, this page should explain it ok :)

https://www.mathsisfun.com/data/standard-deviation-formulas.html

@@ What is Happening?  [Wrap4]    SAt 24/10/15   Sydney, Australia Time 10:15pm   ♪ ♫

Hi,

Not much happened today but here are a couple of interest posts  

1) Mathematics  joke     

Sat    24/10/15

1) Mathematics  joke     

     http://web2.0calc.com/questions/get-it_1#r3

2)  Intergration hiccup       Thanks Alan.

     http://web2.0calc.com/questions/help-with-finding-the-integer-of-x-x-2-2