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- Опубликовано: 4 окт 2024
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a similar problem was on my precalculus final:
10^sin^2 (x) + 10^cos^2 (x) = 11
Is it possible for a six to be 0?, because that's one of the answers that I get, idk if I did something wrong =,C
@iampreyan
It’s sin =0 and sin=+-1 right?
You could also use AM >= GM
yeah its way faster but if its not 6 then things will be different
Can you explain it?
@@Orilliansapply am gm on 9^sin^2x and 9^cos^2x, you get 9^sin^2x + 9^cos^2x >=6. equality holds when the terms are the same, so sin^2x=cos^2x
It's been 2 years since I visited your channel. Good to hear you again Bri.
You sound different. All good?
All good :)
❤️
❤
Different mic?
This one really reminds me of an engineering problem I had in a statics class. I might make a video about that one.
You can actually also solve this using hyperbolic cosine:
9^(sin^2(x)) + 9^(cos^2(x)) = 6
3^(2sin^2(x))/(3^1) + 3^(2cos^2(x))/(3^1)= 2
3^(2sin^2(x)-1)+3^(2cos^2(x)-1) = 2
3^(-2cos(2x))+3^(cos(2x)) = 2
Now let u = 2cos(2x)
(3^(u) + 3(-u))/2 = 1
Change of base into base e
(e^(uln(3)) + e^(-uln(3)))/2 = 1
cosh(uln(3)) = 1
uln(3) = 0
cos(2x)=0
x = (pi*n)/4 +- 2n
Can you make a video on how to calculate cos(i) please
Amazing - thanks!
What about logarithms bro
So I found another cool way to solve this one, not sure if it's 100% mathetically correct though.
First put 9 to the front on the left side, you'll see that it becomes 9(9^(sinx)^2-1 + 9^(cosx)^2-1)
Now use the base formule of trig to see that they they then result back to the base equation from before. We can now see that we can infinitely many times put 9 to the front while keeping the 9^(sinx)^2 + 9^(cosx)^2
Knowing that we can keep dividing both sides by 9 infinitely many times so we can say that 9^(sinx)^2 + 9^(cosx)^2=0
Subtract 9^(cosx)^2 from both sides, then take the 9 log of both and you'll find the simple equation sinx^2 = - cosx^2. If you solve this you'll get the same answer.
t=9^(sin^2(x))>0, t+9/t=6, t^2-6t+9=0, (t-3)^2=0, t=3, 9^(sin^2(x))=3, sin^2(x)=1/2, cos(2x)=1-2sin^2(x)=0, 2x=pi/2+pi k, x= pi/4 + pi k / 2 (k is integer).
Is it invsin(sqrt(log_9(5)))?
Please try to integrate (1/(x^5)+1)dx
Let us say cos^2x = 1- sin^2x , then we may form a quadratic , (such that d=o ) and solve it to get sin^2x = 0.5
Ez arithmetic mean is greater than or equal to Geometric mean
bruh I skipped 10 secs, gone from sin to y pie something
I have done it with 16^sin^2 +16^cos^2=10
Very useful
Vvvfg❤🎉😮😮xzglju😅😮😢😢.bg🎉😮
Probably yea
Hi!
we know 0! is 1, 0!! is 1 as empty product but other answer is √(2/π), what about 0!!! ? 0!!! have other answer or only one answer is 1?
Since sine is the derivative of cos you could use that for these calculations
Hey! Can you solve this? Idk the answer though
int(0 to π) sin(x²) dx
If you solve it then I will say thanks
Log?
won't work
Would it work if I apply Log base 9 on all elements to get rid of the 9 and then use the property of (sin x)^2 + (cos x)^2 = 1? 🤔
No man, this property only works for product, here we have a sum
@@gileadedetogni9054 Thanks!
Good guess but we have a sum of exponentials rather than a product of them
(π+2πk)/4
x = (√2)/2
You forgot it's sin(x) at the end, but we came to the same conclusion
@@luffy.bcos x will also be same √2/2
As a 8th grader student I solved this question by seeing 👀 your thumbnail 👌 it was really easy. The answer is 45°
WHAT
Sorry but, sin^2(x) (sin(x))^2
This is moderate level question of jee exam.
Lol yes
@@Matrix_932 what exam are you preparing for if this seems difficult this is literally not even near jee level
Edit: you replied to me or him?
It is actually way wayyy easier than a moderate level jee question , even a 10th grader can do this
Bro I'm from grade 10 this is a basic question in Olympiads
@@ai2657yeah
The question is made from exponential properties from class 9th and trigonometry from 10th
Btw I'm in 10th too...
Bro there is a wrong step.
√Sin²x=√½
=>Sinx=1/√2
=>Sinx=Sin45°
=>x=45
1/sqrt(2) = sqrt(2)/2
No, he just rationalised the denominator
Yeah I agree but isn't that a good way to assume x as 45?
@@pranavtiwari_yt yeah I agree 👍 but bro without rationalizing we can prove x as 45
@@anupamamanna8989I am assuming you haven't learnt the concept of radians yet so I will only tell you that pi/4 radians(45°) is only one of the many solutions
The answer could be 45°,135°,405° and so on