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Hopecantid

Member

Registered: 2008-04-08

Posts: 13

Triginometry Help

I have a question with three problems please, if anyone wouldn't mind.  A detailed response would be preferred if you have time, since I'd like to understand it instead of just getting an answer.    Thank you for any help, it is greatly appreciated! 

1.  Give exact answer in radians.

2. Solve the following equation for

3. Find all values of x

such that

Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 46,406

Re: Triginometry Help

Answer to #1.
pi/6 radians.

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Kurre

Member

Registered: 2006-07-18

Posts: 280

Re: Triginometry Help

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.

Last edited by Kurre (2008-05-14 02:35:01)

Kurre

Member

Registered: 2006-07-18

Posts: 280

Re: Triginometry Help

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
edit: or you can ofcourse answer x=arccos(...), thats the exact answer. Stupid me:/

Last edited by Kurre (2008-05-14 04:51:39)

Hopecantid

Member

Registered: 2008-04-08

Posts: 13

Re: Triginometry Help

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

Kurre

Member

Registered: 2006-07-18

Posts: 280

Re: Triginometry Help

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:

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

orcared

Member

Registered: 2008-07-12

Posts: 6

Re: Triginometry Help

Solution to the first problem:

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