## lilsis76 Group Title suppose sin theta = 1/5 and 0<theat< pi/2 compute cos theta and compute sin 2theta... i have these formulas given but i dont know what to do with them. let me draw them one year ago one year ago

1. lilsis76 Group Title

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2. phi Group Title

i have these formulas given... The expressions you posted are specific examples, not formulas. To answer this question suppose sin x = 1/5 and 0<x< pi/2 (Let me use x, it's easier to type than theta) First, recall the definition of sin: opposite over hypotenuse. 2nd, it is helpful to draw a graph. |dw:1350682184808:dw|

3. phi Group Title

they tell us x is between 0 and pi/2 (90º) so we know x is in the first quadrant. the cos x is adjacent over hypotenuse. we know the hypotenuse (it's 5) use pythagorean theorem to find the bottom leg of the right triangle (that will be the adjacent side we need for the cos) can you do that?

4. lilsis76 Group Title

yes let me try it on a paper.

5. lilsis76 Group Title

a= sqrt 24 -> i can reduce giving me a=sqrt 4 sqrt 6 a= 2sqrt6

6. phi Group Title

yes, use a's value to find $\cos(x)= \frac{2\sqrt{6}}{5}$

7. phi Group Title

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8. lilsis76 Group Title

sorry about that, this thing is being slow, this openstudy site

9. phi Group Title

to find compute sin 2x I remember sin(a+b)= sin(a)cos(b)+cos(a)sin(b) if a and b are the same=x, we can write this as sin(x+x) = sin(x) cos(x) + cos(x) sin(x) sin(2x)= 2 sin(x) cos(x) so replace sin(x) with 1/5 and cos(x) with 2sqrt(6)/5 in that formula and simplify to get sin(2x)

10. phi Group Title

being slow Yes, sometimes it acts up. Some days are better than others.

11. lilsis76 Group Title

okay ill try to find the first one, but how did you get $2\sqrt{6} / 5 ? <--- how did u get the 5?$

12. phi Group Title

the cos x is adjacent over hypotenuse

13. lilsis76 Group Title

how did u get the 5? haha sorry i cant read the other writting either. and i did sin 2x= sin 2(1/5) - that would make it sin 2/5 right?

14. lilsis76 Group Title

okay i see the cos =5 now. is 2/5 right?

15. phi Group Title

no. it is not the simple (otherwise it would not be math) first, what do you have for sin(x) and cos(x)?

16. lilsis76 Group Title

sin x is opp/hyp= 1/5 cos x is adj/hyp= 2sqrt{6}/5

17. phi Group Title

ok, that is good. I repeat my post from above to find compute sin 2x I remember sin(a+b)= sin(a)cos(b)+cos(a)sin(b) if a and b are the same=x, we can write this as sin(x+x) = sin(x) cos(x) + cos(x) sin(x) sin(2x)= 2 sin(x) cos(x) so replace sin(x) with 1/5 and cos(x) with 2sqrt(6)/5 in that formula and simplify to get sin(2x)

18. lilsis76 Group Title

in the sin(a+b)= sin(a)cos(b)+cos(a)sin(b)? like sin( 1/5 + 2sqrt6/5) = sin 1/5 cos 2sqrt6/5 + cos 1/5 sin 2sqrt6/5

19. phi Group Title

no. don't get confused. I was showing that the "sum of angles" formula simplifies to sin(2x)= 2 sin(x) cos(x) For example, if x were 30º, you could use this formula to find the sin(60º) sin(60)= sin(2*30)= 2 sin(30) cos(30) in this case, you are asked to find the sin(2x). You know sin(x) and cos(x). Can you find sin(2x)?

20. phi Group Title

sin x is opp/hyp= 1/5 cos x is adj/hyp= 2sqrt{6}/5 replace sin(x) with 1/5 replace cos(x) with 2sqrt{6}/5 in the formula sin(2x)= 2 sin(x) cos(x)

21. lilsis76 Group Title

okay let me try it. i am trying to find it on my paper

22. lilsis76 Group Title

okay i got this then

23. lilsis76 Group Title

sin 2x = 2 sin 1/5 cos 2sqrt 6 / 5

24. phi Group Title

be careful x is not 1/5 sin(x) is 1/5 try again.

25. lilsis76 Group Title

okay

26. lilsis76 Group Title

sin 2x = 2sin 1/5 cos x

27. lilsis76 Group Title

thats all im understanding

28. phi Group Title

when we say sin x =1/5 and then write, for example cos(x) sin(x) we can replace sin(x) with 1/5 (that is what = means) so cos(x) sin(x) could be written as cos(x) * 1/5 can you use that idea in sin(2x)= 2 sin(x) cos(x) when sin x= 1/5 cos x= 2sqrt{6}/5

29. phi Group Title

in other words, replace sin x with 1/5 and replace cos(x) with 2sqrt(6)/5

30. lilsis76 Group Title

into the formula you gave me? the...sin(2x)= 2 sin(x) cos(x) ? what is that equation for also?

31. phi Group Title

yes, in the formula sin(2x)= 2 sin(x) cos(x) I'll post a video that explains this, so you can watch it when you have time.

32. lilsis76 Group Title

okay, thank you

33. lilsis76 Group Title

like this: 2(1/5) = 2(1/5)(2sqrt6 / 5) ?

34. lilsis76 Group Title

HEYY?! i just realized, did we solve the compute cos theta? was that the 2sqrt6 / 5?

35. phi Group Title

closer. we know sin(x) and cos(x). but we do not know sin(2x) we start with sin(2x)= 2 sin(x) cos(x) replace sin(x) with 1/5 sin(2x)= 2* 1/5 * cos(x) replace cos(x) with 2sqrt6/5 sin(2x)= 2* 1/5 * 2sqrt6/5 notice that sin(2x) is not 2*1/5

36. lilsis76 Group Title

hmmm....okay

37. lilsis76 Group Title

will the 2 (1/5) = 2/5? or is that wrong?

38. lilsis76 Group Title

i ddont know how to mult. fractions.

39. phi Group Title

yes 2*1/5 is 2/5 the answer is $\sin(2\theta)= \frac{4\sqrt{6}}{25}$ (I used theta again just to match the problem)

40. phi Group Title

Here is the first video on trig http://www.khanacademy.org/math/trigonometry/v/basic-trigonometry

41. lilsis76 Group Title

oh....okay, thank you ill watch the video. but one quesiton. why dont we multiply the sqrt by the 2 also?

42. phi Group Title

Here is a list of videos on fractions http://www.khanacademy.org/math/arithmetic/fractions I would watch the first few just to see if they make sense. Later on, if you run into a problem, go to this site and search for a video that explains it.

43. lilsis76 Group Title

okay. thank you so much. imma watch a few of these then get back to the homework.

44. phi Group Title

$\sin(2x)= 2\cdot \frac{1}{5} \cdot \frac{ 2\sqrt{6}}{5}$ to multiply fractions you multiply top*top and bottom*bottom (whole numbers are "tops") $\sin(2x)= \frac{2\cdot 1 \cdot 2\sqrt{6}}{5\cdot 5}$ that simplifies to $\sin(2x)= \frac{4\sqrt{6}}{25}$

45. lilsis76 Group Title

oh...okay now i see it, dang lol i never saw it like that before. well thank you I shall watch the videos and ill be back later