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mitchelsewbaran
 2 years ago
Verify the identity.
cos 4u= cos^2(2u) sin^2 (2u)
mitchelsewbaran
 2 years ago
Verify the identity. cos 4u= cos^2(2u) sin^2 (2u)

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mathslover
 2 years ago
Best ResponseYou've already chosen the best response.2Put 'u' as any angle ... let it be \(\large{\frac{\pi}{2}}\) .

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.2What do you get now as : \(\large{\cos(4(\frac{\pi}{2}))}\) = ?

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.2Can you tell me @mitchelsewbaran

.Sam.
 2 years ago
Best ResponseYou've already chosen the best response.1I would use cos(4u) cos(2u+2u) Then by identity: cos(A+B)=cosAcosBsinAsinB cos(2u)cos(2u)sin(2u)sin(2u) cos^2(2u)sin^2(2u)

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.2Is that verification or proof @.Sam. I think we have to verify and not to prove

mitchelsewbaran
 2 years ago
Best ResponseYou've already chosen the best response.1cos(4(π 2 )) = 1

.Sam.
 2 years ago
Best ResponseYou've already chosen the best response.1If they both equal then that's verifying too

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.2right now solve this : \[\large{\cos^2(2(\frac{\pi}{2})) \sin^2(2(\frac{\pi}{2}))}\]

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.2yes can you tell me how you got that ?

mitchelsewbaran
 2 years ago
Best ResponseYou've already chosen the best response.1I replaced that whole equation with cos(2pi). Taking the cosine of 2pi, I got 1

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.2o.O a small mistake @mitchelsewbaran : see here: \[\large{\cos^2(2(\frac{\pi}{2}))  \sin^2(2(\frac{\pi}{2}))}\] \[\large{\cos^2(\pi)  \sin^2(\pi)}\] \[\large{10= 1}\]

mitchelsewbaran
 2 years ago
Best ResponseYou've already chosen the best response.1oh srry about that

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.2No problem I just corrected you and I hope that you will not repeat that again. Best of luck :)

mitchelsewbaran
 2 years ago
Best ResponseYou've already chosen the best response.1I have only 2 more that I need to verify. I was wondering if you can help?

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0You cannot substitute a single value (or any finite number of values) to PROVE this. If you find one that doesn't work, that would be sufficient to DISprove it.

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.2Did you try yourself first?

mitchelsewbaran
 2 years ago
Best ResponseYou've already chosen the best response.1can u help me verify these last 2?

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.2there is not fight @mitchelsewbaran . sorry if you felt bad. @tkhunny the question is to verify .. and hence we have to put a value and verify it

mitchelsewbaran
 2 years ago
Best ResponseYou've already chosen the best response.1@mathslover ok :)

mitchelsewbaran
 2 years ago
Best ResponseYou've already chosen the best response.1no I didn't feel bad @mathslover

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0@mathslover Demonstrating one or two values is just not what it is asking. It must be verified for ALL values. Since you cannot enter infinitely many values, there must be other methods employed. Demonstrating \(x = \pi\) says nothing of \(x = \sqrt{2}\)

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.2Ok @mitchelsewbaran post the one question of that two there in "ask a question forum" and surely I will help you but the second one will be solved by you and I will check it... if you get any problem in any concept you can ask me.

mitchelsewbaran
 2 years ago
Best ResponseYou've already chosen the best response.1@mathlover u're awesome, bro :)

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.2well I think @tkhunny has a point but still it is not given to verify for all values... it is just to verify.. but if we still go on for a proving method .. then it's surely not verifying

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0No, this is just not the case. "Verify the Identity" has meant to provide a general proof for all possible values in the entire Domain  at least since 1972. I am familiar with no reference in any mathematical text that has ever meant anything else. If only a few values were to be sufficient, the problem statement would say something like this, "Verify a few values and formulate a conjecture on whether or not this statement is generally true." "Verify the Identity" means ALL of them  leaving nothing out.
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