clara1223
  • clara1223
simplify the following, where k represents any integer: cos(2kpi-x)(-sin(2kpi-x)) a) -cosx*sinx b) (1/2)sin(2x) c) cos((1/2)x)^2 d) cos(x)^2*sin(x)^2 e) undefined
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
clara1223
  • clara1223
@Loser66 could you explain how you got that answer please?
dumbcow
  • dumbcow
https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Symmetry.2C_shifts.2C_and_periodicity https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Double-angle.2C_triple-angle.2C_and_half-angle_formulae
mathstudent55
  • mathstudent55
Here is the explanation: Since the sine and cosine are periodic functions with a period of 2pi, sin (2kpi - x) = sin (-x) and cos(2kpi - x) = cos (-x) Ok so far?

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mathstudent55
  • mathstudent55
Then because of even and odd functions, sin (-x) = - sin x and cos (-x) = cos x
mathstudent55
  • mathstudent55
Putting it all together, you get: \((\cos (2k \pi-x)) ((-\sin(2k \pi-x)) \) \(= \cos (-x) \sin (-x)\) \(= - \sin x \cos x\)
clara1223
  • clara1223
@mathstudent55 thanks so much!
dumbcow
  • dumbcow
wait ...... -sin(-x) = sin(x) it equals positive sin(x) cos(x) then use double angle identity for sin ----> sin(2x) = 2sincos ----> sincos = 1/2 sin(2x)
mathstudent55
  • mathstudent55
@dumbcow You're correct. I missed the negative with the sin in the original problem.
mathstudent55
  • mathstudent55
This is how it should read: \((\cos (2k \pi-x)) ((-\sin(2k \pi-x))\) \(= \cos (-x)(- \sin (-x))\) \(= \cos x (-(-\sin x))\)\) \(= \sin x \cos x\) \(=\dfrac{1}{2} \sin 2x\)

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