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Help me to solve this equation
I do my best to solve this one , could any one help me please ,
this is the equation
Cosine 2x =Cosine² x -Sine² x
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Senior Member
That's an identity, true for all arguments. (Infinitely many solutions for x).
http://mathworld.wolfram.com/Double-AngleFormulas.html
If you're asking how to prove it, see equations 8-11 of:
http://mathworld.wolfram.com/Trigono...nFormulas.html
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thankyou for the links but I try in many ways and it doesn't come !!!
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please I need help , there is someone was help me before , I need this help again !
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Banned-ed-er-ing
The equation is solved. If your trying to do it in a calculator, TI-89 for example, then make sure that.
1. Your in radian mode.
2. The typed equation looks like this
cos(2x) = (cos(x))² - (sin(x))²
3. Also, be sure to specify that you only want values between 0>x>360. Otherwise, you will get infinitely many solutions, as previously stated by rachil0.
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Originally Posted by WMLeeBo
The equation is solved. If your trying to do it in a calculator, TI-89 for example, then make sure that.
1. Your in radian mode.
2. The typed equation looks like this
cos(2x) = (cos(x))² - (sin(x))²
3. Also, be sure to specify that you only want values between 0>x>360. Otherwise, you will get infinitely many solutions, as previously stated by rachil0.
I don't mean that ! I want to prove as this member proved to me in this thread
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Banned-ed-er-ing
Ah! Sorry. It seems you found your answer in a double post, which is aginst FK rules.
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Please help solve the system
Please, have a look on this weird system of equations:
a*(sinh(x)-sin(x))+b*(cosh(x)-cos(x))=c*(sinh(y)+sin(y))+d*(cosh(y)+cos(y))
b*(sinh(x)+sin(x))+a*(cosh(x)-cos(x))=-d*(sinh(y)-sin(y))-c*(cosh(y)+cos(y))
a*(sinh(x)+sin(x))+b*(cosh(x)+cos(x))=c*(sinh(y)-sin(y))+d*(cosh(y)-cos(y))
b*(sinh(x)-sin(x))+a*(cosh(x)+cos(x))+d*(sinh(y)+sin(y))+c*(c osh(y)-cos(y))+r*t*[ a*(sinh(x)-sin(x))+b*(cosh(x)-cos(x))]=0
Does anyone knows how to find a,b,c,d? (x,y,r,t are given)
I will be very grateful for any hints!
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