anonymous
  • anonymous
Verify the identity. cos 4u= cos^2(2u)- sin^2 (2u)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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mathslover
  • mathslover
Put 'u' as any angle ... let it be \(\large{\frac{\pi}{2}}\) .
mathslover
  • mathslover
What do you get now as : \(\large{\cos(4(\frac{\pi}{2}))}\) = ?
mathslover
  • mathslover
Can you tell me @mitchelsewbaran

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More answers

.Sam.
  • .Sam.
I would use cos(4u) cos(2u+2u) Then by identity: cos(A+B)=cosAcosB-sinAsinB cos(2u)cos(2u)-sin(2u)sin(2u) cos^2(2u)-sin^2(2u)
mathslover
  • mathslover
Is that verification or proof @.Sam. I think we have to verify and not to prove
anonymous
  • anonymous
cos(4(π 2 )) = 1
.Sam.
  • .Sam.
If they both equal then that's verifying too
mathslover
  • mathslover
right now solve this : \[\large{\cos^2(2(\frac{\pi}{2}))- \sin^2(2(\frac{\pi}{2}))}\]
anonymous
  • anonymous
hold on
mathslover
  • mathslover
take your time
anonymous
  • anonymous
1?
mathslover
  • mathslover
yes can you tell me how you got that ?
anonymous
  • anonymous
I replaced that whole equation with cos(2pi). Taking the cosine of 2pi, I got 1
mathslover
  • mathslover
o.O a small mistake @mitchelsewbaran : see here: \[\large{\cos^2(2(\frac{\pi}{2})) - \sin^2(2(\frac{\pi}{2}))}\] \[\large{\cos^2(\pi) - \sin^2(\pi)}\] \[\large{1-0= 1}\]
anonymous
  • anonymous
oh srry about that
mathslover
  • mathslover
No problem I just corrected you and I hope that you will not repeat that again. Best of luck :)
anonymous
  • anonymous
I have only 2 more that I need to verify. I was wondering if you can help?
tkhunny
  • tkhunny
You 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
  • mathslover
Did you try yourself first?
anonymous
  • anonymous
can u help me verify these last 2?
mathslover
  • mathslover
there 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
anonymous
  • anonymous
@mathslover ok :)
anonymous
  • anonymous
no I didn't feel bad @mathslover
tkhunny
  • tkhunny
@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
  • mathslover
Ok @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.
anonymous
  • anonymous
ok :)
anonymous
  • anonymous
@mathlover u're awesome, bro :)
mathslover
  • mathslover
:)
anonymous
  • anonymous
;)
mathslover
  • mathslover
well 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
  • tkhunny
No, 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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