## anonymous one year ago Could somebody help me with the question: Find all the solutions for the equation: cosx = -1/2 Not looking for straight answers, but verification of the answer after an explanation on how to solve? Thank you!

1. jim_thompson5910

do you have a unit circle with you?

2. anonymous

$$\bf cos(x)=\cfrac{1}{2}\implies cos^{-1}[cos(x)]=cos^{-1}\left( \cfrac{1}{2} \right)\implies x=cos^{-1}\left( \cfrac{1}{2} \right)$$

3. anonymous

hmmm just notice is negative anyway $$\bf cos(x)=-\cfrac{1}{2}\implies cos^{-1}[cos(x)]=cos^{-1}\left( -\cfrac{1}{2} \right)\implies x=cos^{-1}\left( -\cfrac{1}{2} \right)$$

4. anonymous

The whole equation you replied with isn't showing up... @jdoe0001 But, I pulled up a unit circle.

5. jim_thompson5910

look at the points on the unit circle that have an x coordinate of -1/2 what are the angles that correspond to these points?

6. anonymous

240 degrees and 120 degrees? With the radian values being 4pi/3 and 2pi/3 respectively? @jim_thompson5910

7. jim_thompson5910

very good

8. jim_thompson5910

those are 2 of infinitely many angle values x that make cos(x) = -1/2 true

9. jim_thompson5910

add on 360 (in degree mode) or 2pi (radian mode) to get coterminal angles. You can also subtract 360 or 2pi to get other coterminal angles. There is no limit to how much you can add or subtract

10. jim_thompson5910

So if you're in degree mode, then the solution set is x = 120 + 360*n or x = 240 + 360*n where n is an integer

11. anonymous

So with the answer choices of: A. {2pi/3+npi | n = o. +/-1, +/-2, ... } B. {5pi/6+npi | n = o. +/-1, +/-2, ... } C. {5pi/6+2npi, 7pi/6+2npi | n = o. +/-1, +/-2, ... } D. {2pi/3+2npi , 4pi/3+2npi | n = o. +/-1, +/-2, ... } Would the answer be D? @jim_thompson5910

12. jim_thompson5910

yes it is

13. anonymous

And could you help me with 1 more?

14. jim_thompson5910

sure

15. anonymous

Well, three actually. I think I got them correct, but just verifying. 1. For cos94cos37 + sin94sin37 (degrees) would the answer be cos57? 2. For sin8xcosx - cos8xsinx, I got (in terms of sin) sin9x 3. And for cos8xcos2x - sin8xsin2x, I got cos10x Just confused, on those, when to use (cos a + b), (cos a - b), (sin a + b), or (sin a - b)

16. jim_thompson5910

#1 is correct you can use a calculator to get cos(94)*cos(37)+sin(94)*sin(37) = 0.54463903501502 cos(57) = 0.54463903501502 both are equal to the same decimal value. You can also subtract the two values and you'll get 0 or very close to it

17. jim_thompson5910

#2 is incorrect you use sin(A-B) = sin(A)cos(B) - cos(A)sin(B)

18. jim_thompson5910

#3 is correct cos(A-B) = cos(A)cos(B) + sin(A)sin(B)

19. anonymous

So for#2, instead of sin9x, would it be sin7x?

20. jim_thompson5910

yes

21. anonymous

Can you explain when to use which identity? Like when to use the sin difference or sum, vs. the cos difference or sum? And when to use - vs +?

22. jim_thompson5910

when you have the two 'cos' terms together, like with cos(A)cos(B) + sin(A)sin(B), you use the cos(A+B) or cos(A-B) identity

23. jim_thompson5910

whatever symbol is between the cos(A)cos(B) and sin(A)sin(B) is going to be the opposite inside the cosine on the left side

24. jim_thompson5910

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25. jim_thompson5910

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26. jim_thompson5910

and if you have sin mixed with cosine, like sin(2x)cos(x) + cos(x)sin(2x), then you'll use sin(A+B) or sin(A-B)

27. jim_thompson5910

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28. jim_thompson5910

It comes down to recognizing the form. That happens with lots of practice and memorization. You can look up the identities on a reference sheet like this one http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf (see the "Sum and Difference Formulas" section on page 2) but sometimes the teacher won't let you have a reference sheet like that. It all depends.