All Answers 358516 Answers

Angle DBA = 44 degrees.

The area of triangle XYZ is 42*sqrt(3).

y = 2x^2 - 8x + 5.

The solutions are sqrt(2)*e^(2*pi*i/12), sqrt(2)*e^(2*pi*7*i/12), and sqrt(2)*e^(2*pi*13*i/12).

2x99=198

As shown below what? Can't see picture.

Consider how many beads are there between the two beads. And, try harder next time because you can just list all the cases.

You can simply solve this. The degree is only two.

Thank you!!!

Yes exactly!
You are welcome :).

Do not hesitate to ask any questions when you are confused!

Thank you guest, great help...apreciate!

oohh, i see,

I saw the 16 Sin and 8 Sin as 16 apples and 8 apples, so we are left with 2 apples..(2Sin)...SO, the Sin in itself cancels?....so 16 Sin devided by 8 Sin =2...and NOT 2Sin...I understand now...thank you for your patience and assistance!!...I do appreciate!!!..Thank you!

sarcasm..I really do not need that..I need to understand these things...If you want, I'll just refrain from asking anything here again..

Analysis of sammyadam’s comment indicates moronic incompetence, not sarcasm.

However, if you ...refrain from asking anything here again... that would be WONDERFUL and much APPRECIATED. 

[This isn’t sarcasm, but it may be a little hyperbolic...] 

--. .-

I didn't use any identity till the factorization.

Only after the factorization I replaced 2cos^2(x) -1 with cos(2x)

Hi,

Ok so the question is: \(16sin(x)cos^3(x)-8sin(x)cos(x)\)

Now, if we factor \(8sin(x)cos(x)\), then there should be no sin(x) in either of the terms in the other bracket right?
\(8sin(x)cos(x)[2cos^2(x)-1]\).

To check, expand it again, you will get the question.

But your factorization: \(8sinxcosx(2sinxcos^2x-1)=16sin^2xcos^3x-8sinxcosx\)

So you put an extra sin(x) in the first term. 

Here is a more clarification:

\(16sin(x)cos^3(x)-8sin(x)cos(x) \\ \text{Let us factor sin(x) first only} \\ sin(x)[16cos^3(x)-8cos(x)] \\ \text{Correct so far right?} \\ \text{Now, let's factor 8cos(x)} \\ sin(x)*8cos(x)[2cos^2(x)-1] \\\)

Which is how I got the factorization.

simplify      [16Sin(x) * Cos^3(x)  -  8Sin(x)Cos(x) ]

Results
2 sin(4 x)
8 sin(x) cos(x) (2 cos^2(x) - 1)
4 (1 - 2 sin^2(x)) sin(2 x)
8 sin(x) cos(x) - 16 sin^3(x) cos(x)
16 sin(x) cos^3(x) - 4 sin(2 x)
8 sin(x) cos^3(x) - 8 sin^3(x) cos(x)
i e^(-4 i x) - i e^(4 i x)
16 sin(π/4 - x) sin(x) sin(x + π/4) cos(x)

Note: Again "WolframAlpha" gave all the above results, which are supposed to be equivalent.

Thank you guest,

I'll check that out..stay blessed..

x = 2 π n - sin^(-1)(3/sqrt(13)) and n element Z

x = 2 π n + π + sin^(-1)(3/sqrt(13)) and n element Z

Hi juriemagic: I entered your question into "WolframAlpha" calculating engine and it gave me the above 2 answers. They give step by step solution but you have to subscribe to it in order to be able to see them. I think it costs about $65 US per year.

P.S. They generally use "radians" instead of "degrees" in solving Trig problems.

guest,

please just a little something..I have scrutenized tghe list of identities you gave, but still fail to see how you got your remainder after factorisation:

you've got:

(\(8Sin(x)Cos(x)[2Cos^2(x)-1]\)

I got:

(\(8Sin(x)Cos(x)[2Sin(x)Cos^2(x)-1]\)

what happenend to the 2Sin(x)?