+1124 Hi friends...please help...
\(16Sinx.Cos^3x-8SinxCosx\)
I factored and got:
\(8SinxCosx(2SinxCos^2x-1) \)
stuck..this is so frustrating!!..where can I learn these identities?...There are hundreds of these these identities, how is it at all possible to study and just know them all?..please, I really need some desperate help to assist me in future as well, I can't keep asking the same kind of questions?...
Hi JurieMagic!
Question: \(16sin(x)cos^3(x)-8sin(x)cos(x)\)
When you factored it, you forgot to remove it from the other bracket.
I.e. the correct factorization is:
\(8sin(x)cos(x)[2cos^2(x)-1]\)
Now, we use this identity: \(cos(2x)=cos^2(x)-sin^2(x)=2cos^2(x)-1\)
So:
\(8sin(x)cos(x)(cos(2x))\)
is the answer. However, we can even simplify further:
\(4(2*sin(x)cos(x))*cos(2x)\)
And another identity: \(sin(2x)=2sin(x)cos(x)\)
So:
\(4(sin(2x))cos(2x)\)
But we use this identity again:
\(4sin(2x)cos(2x)=2(2sin(2x)cos(2x))=2sin(4x)\)
I hope this helps, and do not hesitate to ask further questions!
Another thing, yes, there are a lot of identities in trigonometry.
But do not worry about it. It just takes time. The more questions you solve, the more identities you will use, and so you will remember them after using them a lot.
But, here, I will Include a summary of the most common ones:
\(sin^2x+cos^2x=1 \\ sin(2x)=2sin(x)cos(x) \\ cos(2x)=cos^2x-sin^2x=2cos^2x-1=1-2sin^2(x) \\ tan(2x)=\dfrac{2tan(x)}{1-tan^2x} \\ 1+tan^2(x)=sec^2(x) \\ 1+cot^2(x)=csc^2(x) \\ cos^2(x)=\dfrac{1+cos(2x)}{2} \\ sin^2(x)=\dfrac{1-cos(2x)}{2} \\ \)
\(sin(x+y)=sin(x)cos(y)+sin(y)cos(x) \\ cos(x+y)=cos(x)cos(y)-sin(x)sin(y) \\ tan(x+y)=\dfrac{tan(x)+tan(y)}{1-tan(x)tan(y)}\)
(Notice: In the last three, if you let x=y, you get sin(2x), cos(2x), tan(2x) formula).
So by studying these, you will solve most trigonometric questions.
I hope this helps!
+1124 guest,
you had been helpful beyond believe!!...Thank you for your inputs and I will definitely make a copy of this and go study....Thank you very much!
+1124 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)?
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.
I didn't use any identity till the factorization.
Only after the factorization I replaced 2cos^2(x) -1 with cos(2x)
+1124 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!
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.
+118670 Hi Juriemagic,
NICE ANSWER GUEST!
I am about to post this but it is the same as guests. he/she has explained some things better than I have.
Since I have done it I will post it anyway.
there are not really that many to learn.
You learn a couple and then learn to manipulate them.
the main ones are
\(\boxed{\cos^2 x+sin^2x=1\\ Cos(A\pm B)=cosAcosB\mp sinAsinB\qquad \qquad \text{ The signs are opposite}\\ Sin(A\pm B)=sinAcosB\pm sinBcosA\qquad \qquad \text{ The signs are the same}\\ Tan(A\pm B)=\frac{tanA \pm tanB}{1\mp tanAtanB}}\)
There are some easier ones too like tanA = sinA/cosA but they should already be known to you.
\(16Sinx.Cos^3x-8SinxCosx\\ =8sinxcosx(2cos^2x -1)\\ =4(2sinxcosx)*-1(1-cos^2x-cos^2x )\\ =4(sin2x)*-1(sin^2x-cos^2x )\\ =4(sin2x)(cos^2x-sin^2 x )\\ =4(sin2x)(cos2x)\\ =2[2sin2xcos2x]\\ =2sin(4x)\)
