Precalculus trigonometric identities, any help appreciated! Reddit r/homeworkhelp
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- Опубликовано: 24 июн 2024
- A quick tutorial on how to verify the identity (cos(theta)+cot(theta))/(csc(theta)+1)=cos(theta). This problem is from Reddit r/homeworkhelp: / swdzegv92f
24 trig identity problems: • how to simplify trigon...
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I love watching little tasks like this where simple steps make something that appears complex change into something else.
Is that really an identity? At θ = 0, the starting expression is indeterminate, but cosθ = 1. Those are not the same.
I guess they wanted to write it in form of sine and tan but schools do weird things uk
During the proof, the numerator and denominator are both multiplied by sin θ, so it's taken for granted that sin θ ≠ 0 because otherwise you have 0/0, which is indeterminate, and solving that is out of the scope of Grade 11 math. However, if you can solve it, I'd be willing to bet that the expression still simplifies to cos θ.
@@Duraludon884 is it possible to say 0/0=1?
@@vaanivijay6552 0/0 is indeterminate, but the limit of x/x as x approaches 0 is 1. But there are other expressions that approach 0/0, like 2x/x where the the limit is not 1. The limit of 2x/x as x approaches 0 is 2. That is why 0/0 is indeterminate, since it depends on how you get there.
@@vaanivijay6552We use Year instead of Grade in UK
It’s important to keep in mind the domain: sinθ cannot be equal to 0, which means θ cannot be equal to nπ.
Couldn't we have taken cosθ common without multiplying with sinθ/sinθ? It would result in the same answer but faster, wouldn't it?
Yes you are right 👍
Yes you are right, but the method he used is more generalized and yours is faster
had the same way. Bet u too are Indian
That video made my day
thx !
nah, you can factor the cos out since line 2 and then every thing else will devide each other out
Magic 🤯
Sir Can you plss solve this polynomial equation which has fractional powers that is x^{4/3} - 4x^{2} + 4 = 0 ,
Sir can you make a proper step by step video for this problem, I tried many times but stuck and failed.
So respected bprp sir it's my humble request to you to make a dedicated video on the solution of this equation.
Use the substitution u=x^(2/3). You’ll get a cubic in u. This cubic is irreducible, so the only way I know to solve it is the cubic formula.
Wolfram Alpha suggests rewriting as x^{4/3} = 4x^{2} - 4 then cubing both sides to get x^4 = (4x^2 - 4)^3. However, this does not lead to a "clean" solution, so I would expect you copied the problem down incorrectly.
Indians can do this in seclnds in grade 10 lol
As a Grade 11 student in India I really feel I am litterly too much advance for this kind of math problems 😅
Congratulations. There are bigger challenges in math waiting for you.