## anonymous one year ago 5(cos^2)60 + 4(sec^2)30 - (tan^2)45 = x Find x

1. freckles

well cos(60) and sec(30) and tan(45) can all be evaluated using the unit circle

2. freckles
3. anonymous

Use the trig table........

4. freckles

do you need help using that chart above ?

5. anonymous

.... I need to the values like cos = 1/2 and all.......

6. freckles

cos(60)=1/2 that is correct so squaring both sides gives cos^2(60)=1/4

7. freckles

sec(30) means you need to find cos(30) and flip whatever number that is

8. freckles

tan(45) means you need to find sin(45) and cos(45) but both sin(45) and cos(45) are the same so tan(45)=1

9. freckles

$5 \cos^2(60)+4\sec^2(30)-\tan^2(45) \\ 5(\frac{1}{4})+4(\frac{1}{\cos(30)})^2-1$

10. freckles

so you still need to find cos(30)

11. anonymous

4/3 = sec^2(30)

12. freckles

$\cos(30)=\frac{\sqrt{3}}{2} \\ \cos^2(30)=\frac{3}{4} \\ \frac{1}{\cos^2(30)}=\frac{4}{3} \\ \sec^2(30)=\frac{4}{3}$ that is right so far you have this now: $5(\frac{1}{4})+4(\frac{4}{3})-1$

13. freckles

that can be simplify just be doing a few multiplications and additions(subtractions)

14. anonymous

$\frac{ 5 }{ 4 } +\frac{ 16 }{ 3 } - 1$

15. freckles

right so far

16. anonymous

shall I take LCM?

17. freckles

yes which the lcm(4,3,1)=12

18. freckles

$\frac{5}{4}+\frac{16}{3}-\frac{1}{1} \\ \frac{5(3)}{4(3)}+\frac{16(4)}{3(4)}-\frac{1(12)}{1(12)} \\ \frac{5(3)}{12}+\frac{16(4)}{12}-\frac{12}{12} \\ \frac{5(3)+16(4)-12}{12}$

19. anonymous

67/12

20. freckles

looks awesome!

21. freckles

and totally correct :)

22. anonymous

23. freckles

Sure. I can take a stab at it.

24. triciaal

@freckles would using identities just take longer?

25. freckles

@triciaal if you want to try something with identities you can

26. anonymous

$\frac{ cotA-1 }{ 2 -\sec^2(A) } = \frac{ \cot A }{ 1 + tanA }$

27. freckles

some people don't need to put in terms of sin and cos but I always do because I remember more identities with sin and cos

28. anonymous

Help me?

29. freckles

$\frac{\frac{\cos(a)}{\sin(a)}-1}{2-\frac{1}{\cos^2(a)}} =\frac{\frac{\cos(a)}{\sin(a)}}{1+\frac{\sin(a)}{\cos(a)}}$ so this is what I would do as a first step second step would be to remove the compound fraction action going on

30. anonymous

I'm suppose to use LHS and get RHS

31. freckles

ok then get rid of the compound fraction action on the the left hand side

32. anonymous

$\frac{ \cos(a) - \sin(a) }{ \sin(a) }$

33. anonymous

and for the denominator 2sin^(a)/cos^2(a)

34. freckles

$\frac{\frac{\cos(a)}{\sin(a)}-1}{2-\frac{1}{\cos^2(a)}} \cdot \frac{\sin(a) \cos^2(a)}{\sin(a)\cos^2(a)} \\ \frac{\cos^3(a)-\cos^2(a)\sin(a)}{2 \cos^2(a)\sin(a)-\sin(a)} \\ \text{ so doing a bit of factoring } \\ \frac{\cos^2(a)(\cos(a)-\sin(a))}{\sin(a)(2 \cos^2(a)-1)} \\ \\ \text{ now recall on the other side we wanted } \cot(a) \\ \text{ on \top } \\ \text{ well we notice } \frac{\cos(a)}{\sin(a)}=\cot(a) \\ \cot(a) \frac{\cos(a)(\cos(a)-\sin(a))}{2 \cos^2(a)-1} \\$ now somehow we have to show that $\frac{\cos(a)(\cos(a)-\sin(a))}{2 \cos^2(a)-1}=\frac{1}{1+\tan(a)}$ any ideas @BlackDranzer

35. freckles

if not read this if you want spoilers: $\frac{\cos(a)(\cos(a)-\sin(a))}{2 \cos^2(a)-1} \cdot \frac{\cos(a)+\sin(a)}{\cos(a)+\sin(a)} \\ \frac{\cos(a)(\cos^2(a)-\sin^2(a))}{(2 \cos^2(a)-1)(\cos(a)+\sin(a))} \\ \text{ well recall double angle identity for } cosine \\ \frac{ \cos(a)}{\cos(a)+\sin(a)}$ now divide both top and bottom by cos(a)

36. freckles

that is I used $\cos(2a)=\cos^2(a)-\sin^2(a) \\ \text{ and also } \cos(2a)=2\cos^2(a)-1$

37. freckles

yep and we can bring down that cot(a) factor we had from earlier (though I normally write my fractions like this 1/(1+tan(a)) )

38. freckles

so the key in our proof was the use of double angle identity for cosine

39. anonymous

Nice.. Thanks... I need help in quadratic equations too...

40. freckles

ok let's see it

41. freckles

hey @triciaal did you come up with a way using identities for the first one ?

42. freckles

I honestly haven't given it a try

43. anonymous

An aero plane flying at 1 km horizontally above the ground make an angle of elevation 60 after 10 seconds its 30 find the speed of the aerplane in km/hr

44. freckles

wait what does that mean what is the angle of elevation 60 or 30?

45. freckles

and what is 30?

46. anonymous

from 60 to 30 in secs from the smae point

47. freckles

oh the angle of elevation changed from 60 to 30 in 10 sec gotcha

48. freckles

|dw:1436680212196:dw|

49. freckles

so to find the speed of the airplane over that 10 sec we will need to find the distance it traveled for both angle of elevations...

50. freckles

|dw:1436680529581:dw|

51. freckles

wait my drawing is off

52. freckles

I will just make two separate drawings |dw:1436680577539:dw| |dw:1436680738212:dw|

53. freckles

We need to find the distance the plane has flown for both pictures

54. freckles

notice that is x for the first drawing and B for the second drawing

55. freckles

then once you find the distances for both pictures you will find the difference of those distances and that resulting distance/10 secs will be the speed of the airplane in that 10 secs you have that change

56. anonymous

How to find the dist?

57. freckles

you can use a trig ratio

58. freckles

for example you are given the height which is 1km (the opposite side of the angle of elevation)

59. freckles

and you are wanting to find the length that airplane has flown for both (which is the adjacent side )

60. freckles

61. freckles

once you computer x and B (don't ask why I called them that; I wasn't thinking too much about the letters to use) $\text{ the speed over that 10 sec change }=\frac{\text{ distance traveled in that 10 sec }}{ 10 \sec} \\ \frac{|x-B|}{10 \sec}$

62. anonymous

Thanks. I will manage the rest...

63. freckles

when you are done with that computation your units should by km/sec but if you don't like how small the number is next to those units you can always covert the sec part to either minutes or hours

64. freckles

oh your question says to put it an hours

65. freckles

lol

66. freckles

just let me know if you need any further help with this one

67. freckles