Another option would be to use the double angle cosine formula in the other direction than the first time , and use cos 2x instead of cos x to get an answer.
Awesome video, I was trying to solve it on my own at first but I got stucked when I got those cosines to the 6th and also I didn't know about the notation you used to give the answer. Thanks a lot for sharing this with us, keep doing that
Great video! I think there might be a quicker way. You can use the formula for the half angle of sine and cosine giving you tan^x=(1-cos2x)/(1+cos2x) and write cos4x=2cos^2x-1. Then you can easily obtain an expression cos2x(cos^2(2x)-cos2x-1)=0 giving you the answer right away.
Thanks for the video!! I did it using a Weierstrass substitution: I put tanx = t. So cos(2x) = (1 - t^2)/(1 + t^2) sin(2x) = 2t/(1 + t^2) This allowed me to do the following: cos(4x) = cos(2*2x) = cos^2(2x) - sin^2(2x) By using the expressions for cos(2x) and sin(2x) in terms of t I got a 6 degree polynomial. t^6 + 3t^4 - 5t^2 + 1 = 0 Then I put t^2 = u. So I got u^3 + 3u^2 - 5u + 1 = 0 u = 1 is clearly a solution and I got the other solutions using quadratic formula. Theb I got the values of x from the values of u. Nice exercise!! :)
I chose the obvious but quite long way, which is finding a formula for cos(4x), then substitutions: tan^2(x)=sin^2(x)/cos^2(x) and then sin^2(x)=1-cos^2(x). When I received an equation only with cos(x) I used another substitution: cos^2(x)=t, which gave me an equation for t of 3rd degree, with a root at 1/2. Second solution to this equation is (1+sqrt(5)/4, the third one is negative, which is not possible for real numbers. So, going back, cos(x)=+/-sqrt(2)/2 or cos(x)=+/-sqrt{(1+sqrt(5))/4}. So the solutions for x are just arccos of these values. I know it's pi/4 for the first one an then every pi/2 of course, but for the second one I can't find any "good" angle, so I just left it like this.
Two equations for finding value of 'x' and 'y' 1) ax + by = C 2) px + qy =C (If "C" is the Real Number as 0,1,2,3......). Can we write or say. "ax+by = px+qy = C" ("ax+by = px+qy") ( Because both are equal to "C"). Is it true or not. please help with example 🙏
Hello SyberMath. Well... Very interesting but... At 4:51 you forgot to put +1, so the original equation must be: 1-(cosx)^2+8(cosx)^6-8(cosx)^4+(cosx)^2+1=0 Or 8(cosx)^6-8(cosx)^4+2=0 Anyway, great video! I learned more about factorization forms👌🏽👌🏽
(tg x)^2=(1/(cos x)^2)-1. cos 4x=2 (cos2x)^2-1. We use cos2x=2(cosx)^2-1. Then is there (1-2(cosx)^2)+(2(cosx)^2-1+1)(2(cosx)^2-1)^2=0. We can use substitution 2(cos x)^2-1=z. Then we have -z+z^3+z^2=0
You could have written otherwise 1-cos2x -----------. + 2Cos²2x-1=0 1+cos2x Multiply both sides on 1+2cos2x At last you get 2cos³2x+2cos²2x-2cos2x=0 Take 2cos2x behind brackets 2cos2x(cos²2x+2cos2x-1)=0 2cos2x=0 And Cos²2x+2cos2x-1=0 And that's it to the anwser
I solved the problem in a different and, to my opinion, easier way, by reworking it into factors of cos(2x): sin^2(x)/cos^2(x)+cos(4x)=0 By rewriting: cos(4x) into 2·cos^2(2x) − 1, sin^2(x) into ½ − ½·cos(2x) and cos^2(x) into ½ + ½·cos(2x) I got: [½ − ½·cos(2x)]/[½ + ½·cos(2x)] + 2·cos^2(2x) − 1 = 0 Multiplying this equation by [½ + ½·cos(2x)], assuming cos(2x) ≠ −1, so x ≠ π/2 + n·π, resulted into: ½ − ½·cos(2x) + [2·cos^2(2x) − 1]·[½ + ½·cos(2x)] = 0 This can be worked into: ½ − ½·cos(2x) − ½·cos(2x) + cos^2(2x) + cos^3(2x) − ½ = 0 Thus simplified into: cos^3(2x) + cos^2(2x) − cos(2x) = 0 And factorized into: cos(2x)·[cos^2(2x) + cos(2x) − 1] = 0 Which easily leads to a number of solutions for 2x: - cos(2x) = 0 => 2x = π/2 + n·π => x = π/4 + n·π/2 - cos(2x) = −½ ± ½·√(1+4) = −½ ± ½·√5 etcetera One more remark: allowing complex values for x will also provide solutions for: - cos(2x) = −½ − ½·√5
Nice... Good that you have used Cos(2X) but not effectively Here is a simple solution.... tan^2X = Sin^2X/Cos^2X = { 1-Cos(2X) }/{1+Cos(2X)} { 1-Cos(2X) }/{1+Cos(2X)} + 2*Cos(2X)*Cos(2X) -1 =0 Cos(2X) = Y (1-Y)/(1+Y) +2Y*Y-1= 0 1-Y + (1+Y)*(2Y*Y-1)=0 1-Y + 2Y*Y-1 + 2Y*Y*Y - Y =0 2Y*Y*Y + 2Y*Y - 2Y =0 2Y*(Y*Y + Y -1) =0 Y =0, (-1+-sqrt(5))/2 |Y| Y =0, (-1+sqrt(5))/2 Cos(2X) = 0, 0.62 2X = 90, 51.8 deg X = 45, 25.9 deg
😀 I've been playing with numerical answers to this for half an hour because I couldn't make it work out right… turned out my dyslexia was transposing 2 numerals in the answers so it _looked_ wrong but was right all the time. Doh! The first answer is guessable because tan 45° = 1 and cos 180° = -1 so 0. The second gives us x = acos[(√(1+√5))/2] = 25.9136461864939° In the first quadrant. Then ¹⁄₂π, π & ³⁄₂π of course for 8 total roots.
At the very end you forgot to sqrt the nominator, so the correct answer should be x=arccos(sqrt(1+sqrt5)/2). A neat solution nonetheless!;)
That's right! Thanks! 😁
@@SyberMath Dp you honestly think anyone would think of doing that to a sixth degree equation though?? Hope you can please respond when you can.
@@leif1075 You mean substituting?
Very nice.also if I use the multiple angle formula it may solve more easily.
Another option would be to use the double angle cosine formula in the other direction than the first time , and use cos 2x instead of cos x to get an answer.
Awesome video, I was trying to solve it on my own at first but I got stucked when I got those cosines to the 6th and also I didn't know about the notation you used to give the answer. Thanks a lot for sharing this with us, keep doing that
Glad it helped!
Really nice problem ! I loved it !
Glad to hear that!
Great video! I think there might be a quicker way. You can use the formula for the half angle of sine and cosine giving you tan^x=(1-cos2x)/(1+cos2x) and write cos4x=2cos^2x-1. Then you can easily obtain an expression cos2x(cos^2(2x)-cos2x-1)=0 giving you the answer right away.
That's a great idea!
Exactly in the first place I have used the same approach.
I gave my own comment before having seen yours. It is exactly the same.
Isn't it cos2x (cos^2(2x) + cos2x-1) = 0 ???
Thanks for the video!! I did it using a Weierstrass substitution:
I put tanx = t.
So cos(2x) = (1 - t^2)/(1 + t^2)
sin(2x) = 2t/(1 + t^2)
This allowed me to do the following:
cos(4x) = cos(2*2x)
= cos^2(2x) - sin^2(2x)
By using the expressions for cos(2x) and sin(2x) in terms of t I got a 6 degree polynomial.
t^6 + 3t^4 - 5t^2 + 1 = 0
Then I put t^2 = u.
So I got u^3 + 3u^2 - 5u + 1 = 0
u = 1 is clearly a solution and I got the other solutions using quadratic formula.
Theb I got the values of x from the values of u.
Nice exercise!! :)
You're very welcome! Good thinking!
I chose the obvious but quite long way, which is finding a formula for cos(4x), then substitutions: tan^2(x)=sin^2(x)/cos^2(x) and then sin^2(x)=1-cos^2(x). When I received an equation only with cos(x) I used another substitution: cos^2(x)=t, which gave me an equation for t of 3rd degree, with a root at 1/2. Second solution to this equation is (1+sqrt(5)/4, the third one is negative, which is not possible for real numbers. So, going back, cos(x)=+/-sqrt(2)/2 or cos(x)=+/-sqrt{(1+sqrt(5))/4}. So the solutions for x are just arccos of these values. I know it's pi/4 for the first one an then every pi/2 of course, but for the second one I can't find any "good" angle, so I just left it like this.
Nice!
Thank you, I appreciate your work❤
You are so welcome
Gagaxara gmertma akaki
I hope you're doing well, Akaki! 😊
@@SyberMath
ჰეჰ მაგარი ამოცანაა, მართლაც გაგახარა ღმერთმა. მაგრა გამიხარდა ქართული სახელის გაგონება❤️❤️
Hi please I need your help with one trigonometry proving
Two equations for finding value of 'x' and 'y'
1) ax + by = C
2) px + qy =C
(If "C" is the Real Number as 0,1,2,3......).
Can we write or say.
"ax+by = px+qy = C"
("ax+by = px+qy") ( Because both are equal to "C").
Is it true or not. please help with example 🙏
THANKS PROFESOR!!!, VERY INTERESTING!!!!!
Glad you liked it!
Hello SyberMath. Well... Very interesting but... At 4:51 you forgot to put +1, so the original equation must be:
1-(cosx)^2+8(cosx)^6-8(cosx)^4+(cosx)^2+1=0
Or
8(cosx)^6-8(cosx)^4+2=0
Anyway, great video! I learned more about factorization forms👌🏽👌🏽
Yes, you are right
Solving it, I found that:
X1=45°
X2=135°
X3=25.915°
X4=154.085°
(tg x)^2=(1/(cos x)^2)-1. cos 4x=2 (cos2x)^2-1. We use cos2x=2(cosx)^2-1. Then is there (1-2(cosx)^2)+(2(cosx)^2-1+1)(2(cosx)^2-1)^2=0. We can use substitution 2(cos x)^2-1=z. Then we have -z+z^3+z^2=0
What software are you using? Thank you
Notability, screen recording and video zoom
What happens to (1-sqrt(5))/2 ?
Great 👍.. Keep feeding our intellectual appetite. 💐
Thank you, I will!
You are great sir.tx
Thank you! Trying my best
WOW THIS IS AMAZING THANKS!
No problem! Thank you!
If you substitute c^2=u, why do you get a cubic instead of 4th power, or am I missing something?
c^6
c^2 = u
(c^2)^3 = c^2 * c^2 *c^2
= c^(2*3)
When pulling Down the Power, you Divide, not Subtract.
2*3 =6
At the beginning tan*2in terms of sec2 which is 1/ cos2
very cool problem
Thank you!
You could have written otherwise
1-cos2x
-----------. + 2Cos²2x-1=0
1+cos2x
Multiply both sides on 1+2cos2x
At last you get
2cos³2x+2cos²2x-2cos2x=0
Take 2cos2x behind brackets
2cos2x(cos²2x+2cos2x-1)=0
2cos2x=0
And
Cos²2x+2cos2x-1=0
And that's it to the anwser
Shows how actual triangle make it possible ?🕉🕉🕉🕉
In the first step itself, taking cos^2 up will make the equation, a cubic equation in cos(2x)
yess
This value is approximately 26 degrees , so there is no nice value and we need arc cos function
Yes!
I solved the problem in a different and, to my opinion, easier way, by reworking it into factors of cos(2x):
sin^2(x)/cos^2(x)+cos(4x)=0
By rewriting:
cos(4x) into 2·cos^2(2x) − 1,
sin^2(x) into ½ − ½·cos(2x) and
cos^2(x) into ½ + ½·cos(2x)
I got:
[½ − ½·cos(2x)]/[½ + ½·cos(2x)] + 2·cos^2(2x) − 1 = 0
Multiplying this equation by [½ + ½·cos(2x)], assuming cos(2x) ≠ −1, so x ≠ π/2 + n·π, resulted into:
½ − ½·cos(2x) + [2·cos^2(2x) − 1]·[½ + ½·cos(2x)] = 0
This can be worked into:
½ − ½·cos(2x) − ½·cos(2x) + cos^2(2x) + cos^3(2x) − ½ = 0
Thus simplified into:
cos^3(2x) + cos^2(2x) − cos(2x) = 0
And factorized into:
cos(2x)·[cos^2(2x) + cos(2x) − 1] = 0
Which easily leads to a number of solutions for 2x:
- cos(2x) = 0 => 2x = π/2 + n·π => x = π/4 + n·π/2
- cos(2x) = −½ ± ½·√(1+4) = −½ ± ½·√5 etcetera
One more remark: allowing complex values for x will also provide solutions for:
- cos(2x) = −½ − ½·√5
That's neat!
It is easyier, if you solve for cos(2x) first. tan(x)^2 = (1-cos(x)^2)/cos(x)^2 = (1-cos(2x))/(1+cos(2x))
cos(4x) = 2* cos(2x)^2-1
cos(2x) =y
2*y^2-1+(1-y)/(1+y) = 0
2*y^3+2*y^2-2*y= 0
1)y=0 =cos(2x) =2*cos(x)^2-1 => cos(x)=+/- sqrt(2)/2
2) y^2+y-1 = 0; y=(-1 + sqrt(5))/2 [ (-1-sqrt(5))/2 is smaller then -1]
2*cos(x)^2-1 = (-1+sqrt(5))/2 => cos(x) = +/- sqrt(1+sqrt(5))/2
A fly flies by while doing this ...... and you got it all wrong.🥴
?
@@SyberMath that means it takes a hell of focus and concentration to get it right ...... even if you choose the right variable change .
@@SousouCell oh ok! 😁
👍👍
I was ended at 8u³-8u²+1=0 and let wolfram solve the rest 😄
Great job anyways !
Thanks!
Nice...
Good that you have used Cos(2X) but not effectively
Here is a simple solution....
tan^2X = Sin^2X/Cos^2X = { 1-Cos(2X) }/{1+Cos(2X)}
{ 1-Cos(2X) }/{1+Cos(2X)} + 2*Cos(2X)*Cos(2X) -1 =0
Cos(2X) = Y
(1-Y)/(1+Y) +2Y*Y-1= 0
1-Y + (1+Y)*(2Y*Y-1)=0
1-Y + 2Y*Y-1 + 2Y*Y*Y - Y =0
2Y*Y*Y + 2Y*Y - 2Y =0
2Y*(Y*Y + Y -1) =0
Y =0, (-1+-sqrt(5))/2
|Y| Y =0, (-1+sqrt(5))/2
Cos(2X) = 0, 0.62
2X = 90, 51.8 deg
X = 45, 25.9 deg
Hello
Hi!
@@SyberMath how are you i am fine.
[tan(x)]^2 + cos(4x)
= [tan(x)]^2+1-8[sin(x)]^2[cos(x)]^2
= 1+1-8(4/16)
=2-2
=0
Should be consider x=[n(45degree)]
😀 I've been playing with numerical answers to this for half an hour because I couldn't make it work out right… turned out my dyslexia was transposing 2 numerals in the answers so it _looked_ wrong but was right all the time. Doh!
The first answer is guessable because tan 45° = 1 and cos 180° = -1 so 0. The second gives us x = acos[(√(1+√5))/2] = 25.9136461864939° In the first quadrant. Then ¹⁄₂π, π & ³⁄₂π of course for 8 total roots.