juriemagic

I have here something that shows that (Sin(x)Cos(x)) can be written as \({1 \over2}(2SinxCosx)\)

How do they get to that?..please?

Tiggsy,

Yes, I see,

The top would then be

\(Cos(72-24)\)

but the bottom part...I really do not see how to re-write that..

Oh yes, I see....Thanks a mill Alan

Hi Alan,

I see, so I then say 

\(Cos22=Cos2(11)\)

\(Cos22=1-2Sin^211\)

\(\sqrt{1-k^2}=1-2Sin^211\)

\(2Sin^211=1-\sqrt{1-k^2}\)

\(Sin11=\sqrt{1-\sqrt{1-k^2} \over2}\)

I think this might be it?

Hi Alan,

Yes, I worked on it and kinda got it right, thank you klindly, however, maybe just one that I do not see how to go about is to calculate Sin11?...Must I devide Sin22 by 2 and then proceed, in which case I'm not sure how to?..or is it an entirely different approach?..please if you don't mind..

Oh my goodness, yes off course, thanks a million...I also now finally understand this..

Hi Tiggsy

okay, you gave for odds we have (3*3*1*3) =27..agrre

for odd/even, you calculated (3*3*1*2) = 18...agree

I have calculated evens....and got (3*2*1*1) = 6

and even/odd is (3*4*1*2) = 24

everything added together is 75?

Hi guest,

Thanks for your input, I'm looking at this now...no, the denominator is 840,Yep sorry, the 1st question (Which I did not post here), gave a different scenario, saying that a lock code must be even numbered, and may not contain digits 0 and 1. Digits may not be repeated. Calculate how many possible combinations are there to open the lock...This question here, is question 2, the second part...sorry about the confusion. So, we originally have 10 numbers. 0 an 1 may not be used, so we have 8 numbers, however the code must be even numbered, so we need to reserve one of the evens for the 4th spot, that leaves 7 numbers available... 

First spot is one of 7 numbers, 2nd spot one of 6 numbers, 3rd spot one of 5 numbers and 4th spot, one of 4 numbers..

Thanks for your ijput, I'm looking at that now

Hi Tiggsy,

Finally somebody responds...thank you. I get what you say about how you calculate things...it really makes a lot of sense. Thank you so much. I will go ahead and try to do the last one....I will let you know..Thanks

The thing here is, I do not understand the calculations..