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#151 Re: Help Me ! » Trig Equation » 2008-05-14 08:13:06

Hopecantid wrote:

Is that right?

Almost. neither of sec and tan are defined for 90°, so that one doesnt work.

#152 Re: Help Me ! » Triginometry Help » 2008-05-14 06:53:12

Hopecantid wrote:

I'm sorry, could you please explain number three more in depth?  I'm still confused about it.  sad

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:


one solution is cosx=0, but the original expression isnt defined for that. So dividing by cosx, and then replacing cos²x with z we get:
16z²-20z+5=0
this is a quadratic with solutions:

and since cos²x=z, we get

#153 Re: Help Me ! » Triginometry Help » 2008-05-14 04:08:54

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:


So now I guess you anyway need a calculator to solve for x, so you could just have done that in the beginning smile
edit: or you can ofcourse answer x=arccos(...), thats the exact answer. Stupid me:/

#154 Re: Help Me ! » Triginometry Help » 2008-05-14 02:18:11

2. Solve the following equation for


we have that tanx=sinx/cosx, cotx=cosx/sinx. the expression is not defined for sinx=0, cosx=0, cotx+1=0, so assume none of these are zero. Multiply both sides with 1+cotx yields:

multiply both sides with cosx yields, since tanx=sinx/cosx:




which holds when sin(x)=-cos(x), but then 1+cotx=0, which doesnt work in the original expression.
So it must hold when cosx=1/2->x=pi/3, x=5pi/3.

#155 Re: Help Me ! » function » 2008-04-30 04:12:06

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? smile didnt know for example that differentiable at a single point, could be extended to all points, i think i need to repeat this

#156 Re: Help Me ! » Sigma, Help! » 2008-04-24 01:04:41

uhm, but just squaring the x^2 formula doesnt work..?

#158 Re: Help Me ! » Torque » 2008-04-17 01:27:21

Identity wrote:

[i]

Where did you get the right side from? there is no force mg at that point:/

#159 Re: Puzzles and Games » 2,4,8, what... » 2008-04-15 03:12:18

to bump this thread, here is another sequence: tongue

#160 Re: Help Me ! » Discrete Maths Proves » 2008-04-13 07:44:05

Well if b divides a+2, and b divides a, what can you say about the 2 in (a+2)?

#161 Re: Help Me ! » Discrete Math Proves » 2008-04-12 01:06:30

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)

#164 Re: Help Me ! » Derivative of an inverse function » 2008-04-08 06:52:29

but f(x)=3x²-x  does not have an inverse, or? hmm

#165 Re: Help Me ! » inequality » 2008-04-07 05:31:32

that was cool jane, but you can also do it really easy with jensen's inequality:
the function

is strictly convex for x>0. thus we have

#166 Re: Help Me ! » Surds » 2008-04-06 04:20:01


yea that was neat smile

luca, the answer is NOT ±(√8 + 3), square roots are always positive

#167 Re: Help Me ! » How to find the limit of a sum of a series? » 2008-03-23 04:21:20

hint: what is the sum of two following terms, ie term n and term n+1?

#169 Re: Exercises » Determine all real x » 2008-03-23 01:43:35

assume that x+1/x is an integer. Squaring gives:


which also is an integer
now assume that

are both integers
the following product is an integer:

therefor, x^(n+1) + x^(-n-1) is an integer and by the induction principle it holds for all n.
Hence all x that satisifies the condition is the ones that satisfy that x+1/x is an integer, which implies that there exists infinite many reals satisfying the problems condition.

ps. how do i do the sign, that shows that an element belongs to a set in mathcode?

#170 Re: Exercises » Determine all triples » 2008-03-23 01:12:59

mutliplying all three equations yields:



if x,y,z is not 0

but we also have yz=x so




so all triples are the ones mathsy named

#171 Re: Help Me ! » Help Solving Logarithmic Equations » 2008-03-21 09:45:34

Daniel123 wrote:

Rather than doing all that...

uhm well yea but i wanted to be pedagogic smile

#172 Re: Help Me ! » Help Solving Logarithmic Equations » 2008-03-21 00:23:04

hmm lets try to find ln(-1)




gives a=0 since |-1|=1


#173 Re: Help Me ! » Trig Proof - I am having trouble taking this further » 2008-03-17 06:51:57

aha, I thought he meant:


smile

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.

#174 Re: Help Me ! » Trig Proof - I am having trouble taking this further » 2008-03-17 02:54:10

I just plugged that into a calculator and it didnt match.. yikes

#175 Re: Help Me ! » solve this logarithm » 2008-03-16 05:53:02

heres a solution:


raise both sides with 9:

9=3^2 so...



we have the identity
which gives us:

substituting
and multiplication with 2y and rearranging gives the equation:

which has the roots 2 and -1.
y=2 gives x=9 which works with the original equation, but
is not defined for log3x=-1 (or if you want complex numbers, the logarithm of an imaginary number does not work)
hence, x=9 is the only solution

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