Nice. You forgot the other root -(2-sqrt3) which after taking ln will give you ln(-1) +ln(2-sqrt3) which at the end gives you (2n+1)pi+ i ln(2+sqrt3) so you can combine this and what you got ( 2npi+i ln(2+sqrt3)) to npi + i ln(2+sqrt3) as wolfram.
But if we go this way.. Cos X=2 sinx=√3i e^ix= 2 + i√3i Or e^ix=( 2-√ 3i).e^i2npi x= 2npi- ln(2-√3i) x= 2npi + ln(2+√3i) Can you please point out where did i miss npi in this way?
@@ashwinsenapati3358 this factorisation in sinX and cosX is not unique. Take cosX=-2, sinX=-√3i. Then e^ix=(2-√3).(-1).e^i2nPI. Since -1=e^iPI. We gte e^i(2n+1)PI.
The answer to the question at the end of the video is the other solution t2 of the quadratic. Similarly to how we handle t1, we write t2 in polar form and solve for x: e^ix = -(2-sqrt(3)) = (2-sqrt(3))*e^(2n+1)pi*i Taking the log of both sides gives ix = ln(2-sqrt(3)) + (2n+1)pi*i Thus x = (2n+1)pi + i*ln(2+sqrt(3)) We could just append this to the set of solutions we found before, but let’s be a bit more clever about it. Notice that the two solutions together give us the following pieces of information: 1. The imaginary part is ln(2+sqrt(3)) 2. The real part is either an even integer multiple of pi (from t1) or an odd integer multiple of pi (from t2). Notice that this just means any integer multiple of pi. So the final set of solutions is x = n*pi + i*ln(2+sqrt(3))
Old wine in new bottles. I think you gave this before, but in a different form. If tan(x) = sqrt(3)/2*i then 1+tan^2(x) = 1- 3/4 =1/4 =sec^2(x) so sec(x)=+/ 1/2 and cos(x) = +- 2. About 4 months ago you gave the problem cos(x)=2. I still like your solution.
You determined value of x for positive value of t. The negative value of t gives e^ix = (2- sqrt3)e^(PI+2kPI)i x=(2k+1)PI + ln(2+sqrt3) This solution covers all odd integer multiples of PI and the other solution covers even multiples. When combined the answer becomes x=nPI + ln(2+sqrt3) which confirms WA result. Since tangent repeats every integer multiple of PI this answer makes sense.
Nice! But if you use hyperbolic functions, it eliminates some of the first steps in your calculations. Use the identity: tan(x + nπ) = -i * tanh(ix). Substituting, we get tanh(ix) = sqrt(3) / 2. The definition of tanh(ix) = (e^(ix) - e^(-ix) ) / (e^(ix) = e^(-ix) ), and the rest of this solution proceeds similar to yours.
I put tanx = sinx/cosx and cross multiply to get 2sinx = i✓3 cosx Multiply by -i and put on LHS ✓3 cosx - 2i sinx = 0 Then convering to Euler didn't really help as the modulus is √7 but the RHS is zero. I think there must be something about the tan identity that I generalised.
I didn't like having e's in both the numerator and denominator so I made a triangle with complex sides and used Pythagoras to get cos(x) = 2. Put in exponential form (Euler) and solve to get same x.
Nice. You forgot the other root -(2-sqrt3) which after taking ln will give you ln(-1) +ln(2-sqrt3) which at the end gives you (2n+1)pi+ i ln(2+sqrt3) so you can combine this and what you got ( 2npi+i ln(2+sqrt3)) to npi + i ln(2+sqrt3) as wolfram.
But if we go this way..
Cos X=2 sinx=√3i
e^ix= 2 + i√3i
Or e^ix=( 2-√ 3i).e^i2npi
x= 2npi- ln(2-√3i)
x= 2npi + ln(2+√3i)
Can you please point out where did i miss npi in this way?
@@ashwinsenapati3358 2+i√3i = (2-√3) There should not be any i left.
@@SunilMeena-do7xn oh i forgot to remove that i but still I'm getting 2npi only through this process.. where exactly am I missing out?
@@ashwinsenapati3358 this factorisation in sinX and cosX is not unique. Take cosX=-2, sinX=-√3i. Then e^ix=(2-√3).(-1).e^i2nPI. Since -1=e^iPI. We gte e^i(2n+1)PI.
@@SunilMeena-do7xnok,so in this way, I'm missing out solutions right?
The answer to the question at the end of the video is the other solution t2 of the quadratic.
Similarly to how we handle t1, we write t2 in polar form and solve for x:
e^ix = -(2-sqrt(3))
= (2-sqrt(3))*e^(2n+1)pi*i
Taking the log of both sides gives
ix = ln(2-sqrt(3)) + (2n+1)pi*i
Thus x = (2n+1)pi + i*ln(2+sqrt(3))
We could just append this to the set of solutions we found before, but let’s be a bit more clever about it. Notice that the two solutions together give us the following pieces of information:
1. The imaginary part is ln(2+sqrt(3))
2. The real part is either an even integer multiple of pi (from t1) or an odd integer multiple of pi (from t2). Notice that this just means any integer multiple of pi.
So the final set of solutions is
x = n*pi + i*ln(2+sqrt(3))
Old wine in new bottles. I think you gave this before, but in a different form.
If tan(x) = sqrt(3)/2*i then 1+tan^2(x) = 1- 3/4 =1/4 =sec^2(x) so sec(x)=+/ 1/2 and cos(x) = +- 2.
About 4 months ago you gave the problem cos(x)=2. I still like your solution.
You determined value of x for positive value of t. The negative value of t gives
e^ix = (2- sqrt3)e^(PI+2kPI)i
x=(2k+1)PI + ln(2+sqrt3)
This solution covers all odd integer multiples of PI and the other solution covers even multiples. When combined the answer becomes
x=nPI + ln(2+sqrt3) which confirms WA result.
Since tangent repeats every integer multiple of PI this answer makes sense.
Another potential method is to calculate the principal value of x
If tanx = tana where -PI/2
Fantastic question...💯💯💯.
This equation is one branch of the equation (secX)^2=1/4 and then you get all the solutions
It turns up to the definition of tangent function. What domin and range it should be? 😏😏😏😏😏😏
Nice! But if you use hyperbolic functions, it eliminates some of the first steps in your calculations. Use the identity: tan(x + nπ) = -i * tanh(ix). Substituting, we get tanh(ix) = sqrt(3) / 2. The definition of tanh(ix) = (e^(ix) - e^(-ix) ) / (e^(ix) = e^(-ix) ), and the rest of this solution proceeds similar to yours.
Sir could you please make a proper video on finding out arc length (curve length) of sin function between 0 and 2π
I put tanx = sinx/cosx and cross multiply to get
2sinx = i✓3 cosx
Multiply by -i and put on LHS
✓3 cosx - 2i sinx = 0
Then convering to Euler didn't really help as the modulus is √7 but the RHS is zero.
I think there must be something about the tan identity that I generalised.
I didn't like having e's in both the numerator and denominator so I made a triangle with complex sides and used Pythagoras to get cos(x) = 2. Put in exponential form (Euler) and solve to get same x.
Try this
Evaluate Cos(10 degree)
x = πn - i*ln(2 - √3)
Thanks for an other video master...
Try these problems
Sec(x)=2i
Cot(x)=2i
Can't help thinking that being able to do some maths doesn't make it meaningful. Tan is a function that turns an angle into a ratio.
I did it with a different method ua-cam.com/video/oZxcOMu1v9c/v-deo.html
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