logbase2(sin2x)+logbase2(cos2x) =-1

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Prime Newtons

Prime Newtons

Күн бұрын

Пікірлер: 19
@MathFromAlphaToOmega
@MathFromAlphaToOmega 10 ай бұрын
There's a slight issue with the final answer: Some of those angles will cause sin(2x) and cos(2x) to be -sqrt(2)/2, making the logs undefined in the original equation. To avoid that, we need to take n to be even.
@m.h.6470
@m.h.6470 10 ай бұрын
Not according to Wolfram Alpha...
@dirklutz2818
@dirklutz2818 10 ай бұрын
@@m.h.6470 if n=1 then x=5/8*pi, 2x=10/8*pi and sin(2x)=-sqrt(2)/2. The log then is undefined!
@m.h.6470
@m.h.6470 10 ай бұрын
@@dirklutz2818 So you are saying, that Wolfram Alpha - the best calculation engine out there - gets it wrong, while you are right? Sure...
@vitowidjojo7038
@vitowidjojo7038 10 ай бұрын
​@@m.h.6470 is this because it involves a... complex math? Like how sine are always between -1 and 1 i reals, but can go beyond that on a complex plane. Does log have a complex plane for negatives?
@polktech
@polktech 10 ай бұрын
@@m.h.6470but wolfram alpha literally gives a different result? namely: "x = pi*n + pi/8 , n element Z" which is exactly equivalent to the video solution if n is always even (just looking at the function in geogebra can also reveal this, the function is not defined in for example pi/8+pi/2)
@anghelcobo3285
@anghelcobo3285 10 ай бұрын
Amazing, but in the original equation some values of x will cause log to be undefined so te first step is to find the domain.
@m.h.6470
@m.h.6470 10 ай бұрын
If you give the original equation into Wolfram Alpha, it gives the same exact answer. So without any domain.
@Michaeladjei001
@Michaeladjei001 10 ай бұрын
Wow...such a smooth teaching 😅❤🎉
@kingbeauregard
@kingbeauregard 10 ай бұрын
I was very happy to see the 2*pi*n. A lot of people would have felt that the principal value was sufficient, and It often is; but if we're being complete, we need to allow for all the answers.
@mab9316
@mab9316 23 күн бұрын
It's worth mentioning that it is better to define the domain of the solutions before starting to resolve the equation. In this case: x has to be in 0 to pi/4 +2n.pi
@michaelbaum6796
@michaelbaum6796 10 ай бұрын
Great - very tricky👍
@KaivalyaChess
@KaivalyaChess Ай бұрын
sir you are great
@suvamsugyansahoo
@suvamsugyansahoo 10 ай бұрын
Please do a video on random variables
@justabunga1
@justabunga1 7 ай бұрын
The answer is close but not exactly because if you put in n=1, then the sine of a negative number is negative, but the log of a negative number is undefined. Try n=2, then the answer is defined. Try n=3 and n=4. With respect to those answers, n=3 doesn't work, but n=4 works. You can see that the even integers work. What this tells us is that you have to double the 4n, which is 8n. Therefore, the answer should be (8n+1)π/8 where n is an integer.
@Ron_DeForest
@Ron_DeForest 10 ай бұрын
What happened to the 2 in the equation 2sin2x cos2x = 1/2*2? You went to sin4x but left out the first 2. Did I miss something?
@keithrobinson2941
@keithrobinson2941 10 ай бұрын
He's applying the double angle identity. 2sinxcosx=sin2x (the general equation for this is shown at the bottom of the blackboard), but in today's question, "x" is "2x." Hence, we get sin 4x. See it now?
@ВалерийПигулевский-ъ3ч
@ВалерийПигулевский-ъ3ч 10 ай бұрын
При 5/8pi выражение не существует.
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