27 Replies
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.In 13 on solving the given expression and simplying it u will get a trigonometric ratio and it is just asking the value on 0 , 1st derivative and 2nd derivative
yep thats evident but diff this is a task in itself which i failed at
13 is a very good question it uses what I call the trigno power expansion theorem for even powers
You can write the even powers of cos and sin into multiple angles
Just focus on cos⁶x
Odd powers ke liye kuch dimag lagana hoga
Let me know if you want to know more about how to convert these powers into angles
so this is basically the numerator u have solved
yeah how did u come up with the (cosx)^n formula
Exactly
Give me a few minutes
Basically
Keep this pattern in mind
For cos^nx where n is any even number the series is as seen
For sin^nx we have two cases for multiple of 4 and non multiplies of 4 it has alternating+- signs , whether or not it starts with + or - keep in mind that the final constant term will always be positive
Btw please tell the hint for its derivation @¹¹⁷sos
Use demoivre's theorem
Basically
(Cos nx +i Sin nx)= (Cosx + iSinx)^n
Use bino on rhs and compare real to real and imaginary to imaginary
It will give angle in terms of power for power in terms of angle you will have to do it multiple times and add and subtract in a suitable way
Okay ..
Understood
Thank you @¹¹⁷sos
aight
but your solution seems quite excessive since denominators left too
and then we have to find its derivative
well it has gotten way easier
I don't think there is a better solution to this although I will love it if someone proves me wrong
ill try simple differentiation once since majority of it is cosx, which will become sinx and sin0 is 0 only
Was losing hope finally ek question toh hua
I used sec + tab = 1/(sec - tan)
even iit mandi peeps struggling. thanx a lot brother
Correction= 2Cot2x = cot x - tanx
Also it wasn't a hard question I just wasted a lot of time thinking of the quickest method, this just sounds tedious yet is pretty concise and decent
marking solved. please close the threads that are sorted
+solved @¹¹⁷sos @Real potato
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