Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Verify the identity. cos 4u= cos^2(2u)- sin^2 (2u)

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Put 'u' as any angle ... let it be \(\large{\frac{\pi}{2}}\) .
What do you get now as : \(\large{\cos(4(\frac{\pi}{2}))}\) = ?
Can you tell me @mitchelsewbaran

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question