## clara1223 one year ago 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

1. clara1223

2. dumbcow
3. 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?

4. mathstudent55

Then because of even and odd functions, sin (-x) = - sin x and cos (-x) = cos x

5. mathstudent55

Putting it all together, you get: $$(\cos (2k \pi-x)) ((-\sin(2k \pi-x))$$ $$= \cos (-x) \sin (-x)$$ $$= - \sin x \cos x$$

6. clara1223

@mathstudent55 thanks so much!

7. 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)

8. mathstudent55

@dumbcow You're correct. I missed the negative with the sin in the original problem.

9. 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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