I believe that the best method to solve linear equations is to apply the formulas that transform sin x cosx into tan(x/2) to avoid extraneous solutions or loss of solutions. sinx=2t/(1+t^2) cosx =(1-t^2)/(1+t^2) with t=tan (x/2). You get it right away 3t^2-5t-1=0 then t=2 or t=-1/3 but 0
There is a much easier way. sinx+cosx=1/5, so (sinx+cosx)^2=(sinx)^2+(cosx)^2+2sinxcosx=1/25, [ we know sum of squares of sinx and cosx is always 1] so we will have sinx.cosx=-12/25. Now we have sum and product of sinx and cosx ( 1/5 and -12/25 respectively).sinx and cosx will either 4/5 or -3/5. Because x is between 0 and π sinx must be a positive value, therefore sinx=4/5 and cosx=-3/5 and tanx=-4/3
sin(x)+sin(pi/2-x)=1/5 use the product to sum formula. You will wind up having an unconventional application of the cosine angle addition formula you will have cos(x-pi/4)=1/5sqrt(2) given to you. Let angle a be x-pi/4 for which the cosine is already given to you and angle b be pi/4 for which we already have trig function values for. The cosine will be negative and the only solution for that could be in quadrant 2 and that tells up the sine is positive and the tangent is also negative.
These are fun to watch. After watching several of these videos, I am reminded of my eleventh-grade trig class where we had to do those crazy identities. It also brings back memories of some of my graduate Physics classes where we had problems that took several pages of legal-size paper to solve. The problems expanded out so each step took several lines, then the simplification started and soon the whole mess ended up being a fairly straightforward equation.
Hey man, not tryna be disrespectful or anything but I think the solution you provided is wrong. If you plug in -4/3 in the calculator, sin + cos would be -1/5 but not 1/5, the correct answer should be -3/4. I believe it is because at 0:54 you neglected the case where cosx could be -ve as x is between 0 and pi. For me, I divided everything by cosx and then proceed to square both sides to get a quadratic with tanx (since RHS would have a secx which could be written in terms of tanx after we square it). We would get tanx = -3/4 or tanx= -4/3. Then we observe that when tanx = -4/3, x is between pi/2 and 3pi/4 (with the help of the graph of tanx and the fact that tanx = -1 when x is 3pi/4); but when x is in that domain, both sin and cos are negative. Hence, we have to reject that answer. On the other hand, when tanx= -3/4, x is between 3pi/4 and pi, and cosx is positive in this domain so it is possible to get a positive answer when we add up sin and cos.
You' ve loosed another solution because for cos^2(x)=1-sin^2(x) there are 2 opposite solutions cosx=+ or - square root ... That's why is better to substitute not cosx but sinx because for 0
Sir I think this problem could be solved in a way easier and more simplified way, I mean am I wrong or you've got a purpose for solving this problem this way?❤❤
I believe that the best method to solve linear equations is to apply the formulas that transform sin x cosx into tan(x/2) to avoid extraneous solutions or loss of solutions. sinx=2t/(1+t^2) cosx =(1-t^2)/(1+t^2) with t=tan (x/2). You get it right away 3t^2-5t-1=0 then t=2 or t=-1/3 but 0
Good one, I've not seen those identities used before. Shouldn't the quadratic have -2 for the constant term?
There is a much easier way.
sinx+cosx=1/5, so (sinx+cosx)^2=(sinx)^2+(cosx)^2+2sinxcosx=1/25, [ we know sum of squares of sinx and cosx is always 1] so we will have
sinx.cosx=-12/25. Now we have sum and product of sinx and cosx ( 1/5 and -12/25 respectively).sinx and cosx will either 4/5 or -3/5. Because x is between 0 and π sinx must be a positive value, therefore sinx=4/5 and cosx=-3/5 and tanx=-4/3
sin(x)+sin(pi/2-x)=1/5 use the product to sum formula. You will wind up having an unconventional application of the cosine angle addition formula you will have cos(x-pi/4)=1/5sqrt(2) given to you. Let angle a be x-pi/4 for which the cosine is already given to you and angle b be pi/4 for which we already have trig function values for. The cosine will be negative and the only solution for that could be in quadrant 2 and that tells up the sine is positive and the tangent is also negative.
These are fun to watch.
After watching several of these videos, I am reminded of my eleventh-grade trig class where we had to do those crazy identities. It also brings back memories of some of my graduate Physics classes where we had problems that took several pages of legal-size paper to solve. The problems expanded out so each step took several lines, then the simplification started and soon the whole mess ended up being a fairly straightforward equation.
We have,
sinx + cosx = 1/5
=> (sinx + cosx)/cosx = (1/5)/cosx
=> (sinx/cosx) + 1 = (1/5)secx
=> tanx + 1 = (1/5)secx
=> (tanx + 1)² = (1/25)sec²x
=> tan²x + 2tanx + 1 = (1/25)(1 + tan²x)
=> tan²x + 2tanx + 1 = (1/25) + (1/25)tan²x
=> tan²x - (1/25)tan²x + 2tanx + 1 - (1/25) = 0
=> (24/25)tan²x + 2tanx + (24/25) = 0
=> (12/25)tan²x + tanx + (12/25) = 0
=> 12tan²x + 25tanx + 12 = 0
=> 12tan²x + 16tanx + 9tanx + 12 = 0
=> 4tanx (3tanx + 4) + 3 (3tanx + 4) = 0
=> (3tanx + 4) (4tanx + 3) = 0
=> tanx = - 4/3 or tanx = - 3/4
Hey man, not tryna be disrespectful or anything but I think the solution you provided is wrong. If you plug in -4/3 in the calculator, sin + cos would be -1/5 but not 1/5, the correct answer should be -3/4. I believe it is because at 0:54 you neglected the case where cosx could be -ve as x is between 0 and pi.
For me, I divided everything by cosx and then proceed to square both sides to get a quadratic with tanx (since RHS would have a secx which could be written in terms of tanx after we square it). We would get tanx = -3/4 or tanx= -4/3. Then we observe that when tanx = -4/3, x is between pi/2 and 3pi/4 (with the help of the graph of tanx and the fact that tanx = -1 when x is 3pi/4); but when x is in that domain, both sin and cos are negative. Hence, we have to reject that answer. On the other hand, when tanx= -3/4, x is between 3pi/4 and pi, and cosx is positive in this domain so it is possible to get a positive answer when we add up sin and cos.
3:36 The only change I would make is to divide by 2 and get rid of the 2 in front of the first term.
Thanks, good example.
But interesting is, what's the only possible x: I calculated approximatly 2,21 for x. Approximatly 0,93 for x doesn't work.
Lovely stuff
Thanks and Welcome 🙏❤️🙏
sqrt(cos^2(x))= abs(cos(x)) , not cos(x) such that 0
You' ve loosed another solution because for cos^2(x)=1-sin^2(x) there are 2 opposite solutions cosx=+ or - square root ...
That's why is better to substitute not cosx but sinx because for 0
2*(sinx*cosx)=2/5
Sin2x= 2/5
Draw the triangle and find x on it and boom not wasting 10 minutes
Good. And very clear.
Many thanks ❤️🙏❤️
asnwer=1 isit
I have the angle narrowed to between 120 degrees and 135 degrees. The tangent must therefore be between -sqrt(3) and -1.
Understandable by 13-year-old ones like me😂
Nice you’re very educated then, I’m 18 year old and this exercises are my level
@@VeelgevraagdAboneer Thanks
Same here but 3 years old 😎
@@Usuario459 I don't believe that😏
@@is7728κ
Thanks a lot.
Most welcome!❤️🙏❤️
At 1:04. if x is in the interval (pi/2, pi), does cos(x) = sqrt(1-sin^2(x))?
Thank You Sir.
Thanks, You are most welcome❤️🙏❤️
Sir I think this problem could be solved in a way easier and more simplified way, I mean am I wrong or you've got a purpose for solving this problem this way?❤❤
It's a 3-4-5 triangle, how cool is that?! He did not mention that the answer puts it in the second quadrant.
DIVIDE BOTH SIDES BY COS X AND THEN PROCEED....EASIER
la tine cifra 5 este litera ,,S"" ?
Sin(x) is between -pi and +pi
Coś(x) is between 0 and +pi
no, you are wrong. Sin(x) is between 0 and 1. Coś(x) is between -1 and 1.