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Is that right?
Almost. neither of sec and tan are defined for 90°, so that one doesnt work.
I'm sorry, could you please explain number three more in depth? I'm still confused about it.
The key is to simplify so you get rid of all 2x and 3x.
first, multply with sin2xcos3x to receive:
Cos3xcos2x-sin3xsin2x=0
Do you know about addition formulas for sin and cos? we have that sin(a+b)=sinacosb+cosasinb, and cos(a+b)=cosacosb-sinasinb. We use that for a=x, b=2x and get:
sin(3x)=sin2xcosx+cos2xsinx, cos3x=cos2xcosx-sin2xsinx. Then we have the identities sin2x=2sinxcosx, cos2x=cos²x-sin²x=2cos²x-1.
Using these two identites yields:
cosx(2cos²x-1)²-4sin²xcos³x+2sin²xcosx-4sin²xcos³x+2sin²xcosx-4cos³x+4cos^5x=0
Now, substitute all sin²x with 1-cos²x, and you will get a polynomial in terms of cosx, which after simplifying is:
3. I dont see any "nice" method, I just multiplied with sin2xcis3x, then used addition formula for sin(2x+x) and cos(2x+x) and then fomrula for double angle. Then using the identity sin^2x+cos^2x=1 to express in only cosx, and then dividing with cosx yields a deg 4 polynomial of cosx. Setting cos^2x=z gives a quadratic polynomial, and I got the final answer for cosx to be:
2. Solve the following equation for
I think it is one of the so called cauchys functions, which does not need to be proven. Jane, can you state the exact theorems about how differentiation and continuity relates to another? didnt know for example that differentiable at a single point, could be extended to all points, i think i need to repeat this
uhm, but just squaring the x^2 formula doesnt work..?
use the fact:
[i]
Where did you get the right side from? there is no force mg at that point:/
to bump this thread, here is another sequence:
Well if b divides a+2, and b divides a, what can you say about the 2 in (a+2)?
you dont need induction for the second one, just expand the paranthesis and simplify, and you will end up with an expression that you can easy show that it is divisible by 9 by letting n=3 and n=3k±1. (you can also first change n^3 +(n + 1)^3 + (n+2)^3 to (n - 1)^3 + n^3 + (n+1)^3, which may simplify the expansion)
edit: oh got another one
but f(x)=3x²-x does not have an inverse, or?
that was cool jane, but you can also do it really easy with jensen's inequality:
the function
luca, the answer is NOT ±(√8 + 3), square roots are always positive
hint: what is the sum of two following terms, ie term n and term n+1?
Thanks!
assume that x+1/x is an integer. Squaring gives:
ps. how do i do the sign, that shows that an element belongs to a set in mathcode?
mutliplying all three equations yields:
Rather than doing all that...
uhm well yea but i wanted to be pedagogic
hmm lets try to find ln(-1)
aha, I thought he meant:
Hint: Use addition formula (sin(x+y)=sinxcosy+cosxsiny) and then sin2x=2sinxcosx (for cos, cos(x+y)=cosxcosy-sinxsiny, cos2x=cos^2x-sin^2x). And I think you have some errors in the denominator in the second approach.
I just plugged that into a calculator and it didnt match..
heres a solution: