## anonymous 5 years ago Use an addition or subtraction formula to find the exact value of the expression. sin(285)

1. anonymous

Well, lets assume you know the 'obvious' values of sin - {0, 30, 45, 60, 90} + 360n. How can you use them to make 285?

2. anonymous

3. anonymous

Indeed. Use sin(a+b) with a couple of them and you should be able to work it out. I guess the main problem is how to do it most efficiently.

4. anonymous

5. anonymous

That's where i'm cofused

6. anonymous

Something to consider is that sin (180 - x) = sin x (the best way is probably to add 2 values of sin between 100 and 180

7. anonymous

¬_¬ Sin 150 = Sin 30 Sin 135 = Sin 45

8. anonymous

I do not get it.

9. anonymous

Do you not get why that is true?

10. anonymous

Or do you not get that sin(150 + 135) = sin 285

11. anonymous

both

12. anonymous

Well, if 0 < x < 180 , then sin (180 - x) = sin x Hence the above claim. And 135 + 150 = 285 (clearly) sin(150 + 135) = sin(150)cos(135) + sin(135)cos(150) = sin(30)cos(135) + sin(45)(cos(150) All you need to do is use a SIMILAR rule one the cos values and you're done. I'll leave that to you. NB this is how I'd do it, but there may be a quicker way.

13. anonymous

ok thanks

14. anonymous

Gah, actually I prefer: Sin 285 = sin (285 - 360) = sin(-75) Sin is an odd function => sin(-75) = -sin(75) sin 75 = sin(30+45) = blah. -sin 35 = -blah.

15. anonymous

-90?

16. anonymous

Last line is -sin(75). Where did you get 90 from?

17. anonymous

i have no idea what -sin 35= ?

18. anonymous

It's meant to be -sin(75) , sorry! But sin (75) = sin(45 + 30) = sin(45)cos(30) + sin(30)cos(45) And sin(285) = sin(-75) = -sin(75) = -(sin(45)cos(30) + sin(30)cos(45)) Because sin(-a) = -sin(a)

19. anonymous

20. anonymous

hmm, INew, wouldn't it be easier if you just drew the graph and compute it from the graph? :) it'll be alot easier ^_^

21. anonymous

sin(45) = cos(45) = 1/sqrt(2) sin(30) = 1/2 cos(30) = sqrt(3)/2 $\implies -\sin(75) = -\frac{1}{\sqrt{2}} \cdot \left(\frac{1}{2} + \frac{\sqrt{3}}{2}\right) = -\frac{1+ \sqrt{3}}{2\sqrt{2}} = -\frac{\sqrt{2} + \sqrt{6}}{4} = \sin(285)$

22. anonymous

Easy to compute $-\frac{\sqrt{6} + \sqrt{2}}{4}$ from a graph? You must be pretty badass with graphs.

23. anonymous

lol, yes by drawing the graph. You know turning the lines in circles we have 90 -180 360 -270 he'll easily figure out angle by then :)

24. anonymous

any *

25. anonymous

Just admit you were wrong, please, and I won't have to own you.

26. anonymous

lol, own who?

27. anonymous

the graph is actually easier, never realized it :p

28. anonymous

LOL, do you know how yo use it loco :)?

29. anonymous

The question specifically asks to use the addition formulae, and besides, no 'circle' will give you the angle 285.

30. anonymous

lol, yes it will :) , you've never used it right?

31. anonymous

kind of, ta showed it to us. still iffy on it but practicing.

32. anonymous

besides loco, the information INew has given you also is valuable, work with both ^_^

33. anonymous

thanks guys :)

34. anonymous

np ^_^

35. anonymous

but did you figure it out ? :)

36. anonymous

sstarica, please, draw me a circle which can give the exact value I calculated without, in essence, just using the addition formulae in diagram form.

37. anonymous

I can't draw it here, but you can twirl the line till you get the given angle, then you can figure out which angles to add. Those angles will be easy to compute since they're small, and in the end you can just add them up to find the sin of them ^_^

38. anonymous

stop fighting guys

39. anonymous

Oh, adding angles? Weird, just what I did. BTW drawing is harder than saying 285 = 360 - 75 => we need -sin(75)

40. anonymous

Simplify the trigonometric expression. cos[x]/sec[x] + sin[x]/csc[x]

41. anonymous

Sorry, I thought they were implying they could do the hard part in a circle, not the easy part :/

42. anonymous

I'm not fighting LOL, we're discussing loco ^_^

43. anonymous

1/sec = cos. 1/csc = sin. cos * 1/sec = .... etc

44. anonymous

how about we i mean u guys solve this problem

45. anonymous

you still didn't solve it? .-.

46. anonymous

no, i hate this stuff

47. anonymous

don't say that, you still didn't master it , but you will with practice and it'll be easy on you ^_^

48. anonymous

i've been practicing a lot, but still having trouble :'(

49. anonymous

$\frac{\cos x}{\sec x} = \frac{\cos x}{\frac{1}{\cos x}} = \cos^2x$

50. anonymous

it's alright, keep on practicing and you'll master it :)

51. anonymous

sstarica help?

52. anonymous

hold on :)

53. anonymous

newton it's wrong

54. anonymous

What's wrong? (Before answering, note that I am never wrong)

55. anonymous

sorry, i thought that was the answer

56. anonymous

we all make mistakes Inew :) even I, even the most intelligent, but noone said you're wrong, but your way is complicated.

57. anonymous

try to explain it in a simple way :)

58. anonymous

I was doing one half of it. See if you can apply the same logic to the next part - you might find it simlpifies to.. something nice.

59. anonymous

He meant I am wrong about the second problem he posted in here. I only did half of it

60. anonymous

Yes, my FIRST method was comlpicated, but my second was amazing.

61. anonymous

he posted another question?

62. anonymous

lol, good that you've made it easier ^_^

63. anonymous

Thankfully still no graphs required

64. anonymous

oh that one , LOL LOCO! you've posted it in a different thread =D and I was writing it down

65. anonymous

u can write it here

66. anonymous

nvm it, INew answered it though :)

67. anonymous

it's not right though

68. anonymous

69. anonymous

Alright we have :$\cos(x)/\sec(x) + \sin(x)/\csc(x)$ sec x = 1/cosx and csc x = 1/sin x so: $[\cos(x)/1/\cos(x)] + [\sin(x)/1/\sin(x)]$ simplify and you'll get : $= \cos^2(x) +\sin^2(x)$ $= 1$ Correct me if I'm wrong please ^_^

70. anonymous

you are right :) i got the same answer

71. anonymous

No, you aren't wrong. Unless you count not using \frac{}{} for fractions as wrong (I personally do)

72. anonymous

well done ^_^

73. anonymous

lol, I didn't get you INew

74. anonymous

$\frac{1}{2} \geq 1/2$

75. anonymous

hmm, I can't find the frac() symbol lol >_<

76. anonymous

still don't know how to make that fraction line ^_^"

77. anonymous

last problem.... Simplify the trigonometric expression. cos[x]/ tan[x] + sex[x]

78. anonymous

frac{a}{b} gives$\frac{a}{b}$

79. anonymous

oh, thank you :)

80. anonymous

sex(x) - lolololololol

81. anonymous

LOL

82. anonymous

sec[x], sorry :P

83. anonymous

wish math was that fun

84. anonymous

I general, always write tan as sin/cos , sec as 1/cos etc.

85. anonymous

When trying to simplify things.

86. anonymous

It's alright, we have : $=\cos^2(x)/sinx + 1/sinx$ $= [\cos^2x + 1]/\sin(x)$ $= \sin^2x/sinx = \sin(x)$ ^_^

87. anonymous

ORLY?

88. anonymous

maybe? :) correct me lol

89. anonymous

it's not correct

90. anonymous

$\cos^2x + 1 \not= \sin^2x$

91. anonymous

hmm, where is my mistake

92. anonymous

LOL! right >_< sorry

93. anonymous

For what it's worth, I don't think it simplifies much nicer than how it starts, so is a bad question.

94. anonymous

hmm, solve it :)

95. anonymous

man, u guys are at it.

96. anonymous

LOL, loco we're not fighting =P, hold on :)

97. anonymous

Ugh, it doesn't simplify.. at all. But also sec x = 1/cos(x) ... >_> sorry

98. anonymous

$\sec x + \csc x -\sin x$ .. but I'm tired.

99. anonymous

Loco left is for a new thread. Un coooooooooooool

100. anonymous

i posted it in a new thread also, but u guys were to busy

101. anonymous

2nd one

102. anonymous

103. anonymous

hold on

104. anonymous

Joke, I know it was. Just unsure what the 'expected' form is.

105. anonymous

it's wrong newton

106. anonymous

$=\frac{\cos^2x}{\sin(x)} + 1/cos(x)$$= \frac{\cos(x)(cos^2x+1)}{\sin(x)cos(x)}$$= \frac{2-\sin^2x}{\sin(x)}$$= \frac{2}{\sin(x)} -\sin(x) = 2\csc(x) -\sin(x)$

107. anonymous

I'm not sure though, I was laggin ^_^|"

108. anonymous

just go over it please :)

109. anonymous

nope

110. anonymous

wrong? ._.

111. anonymous

yep

112. anonymous

blah~, I'll try again ^_^

113. anonymous

For the record, loco, my answer (sec(x)+cosec(x) - sin(x)) is right, but probably thr wrong form - it's hideous.

114. anonymous

sec(x) + cos(x)cot(x), maybe?

115. anonymous

don't know, my school's hw program sucks nuts

116. anonymous

i only get 3 tries per problem

117. anonymous

That program is retarded, then. Both of the answers I've given are right, but if the form matters then both could be shown as wrong. FFFFUUUUUUUUUUUUU

118. anonymous

at least it's not timed, i've been doing this since 9pm

119. anonymous

it really pisses me off and the professor is a b**** and doesn't do anything about it

120. anonymous

lol, both of you calm down and just try again ^_^

121. anonymous

$\frac{\cos x}{\tan x} + \sec x = \frac{\cos^2x}{\sin x} + \sec x = \cos x\cot x + \sec x = ?$

122. anonymous

Which can also be written as sec x + csc x - sin x , and many other things.

123. anonymous

124. anonymous

True, she (I think, but maybe he?) does have more fans than me, so that's a good call.

125. anonymous

I'm a she =P

126. anonymous

I'm a she =P

127. anonymous

r u hot? ;)

128. anonymous

just to make sure, what is the question loco? :)

129. anonymous

let me find it

130. anonymous

Simplify the trigonometric expression. cos[x]/ tan[x] + sec[x]

131. anonymous

what on earth? lol what kind of quetion is that? >_>

132. anonymous

idk.... curious loco mind

133. anonymous

Don't worry about loco the sexpest . But in answer to the question of course she is, she does Maths.

134. anonymous

ok... lets continue with math. i don't wanna go to jail

135. anonymous

136. anonymous

I got : cscx -sinx + secx *sigh* not sure though, but that's what I got

137. anonymous

cos^2x + 1 ---------- sinx

138. anonymous

Blexting is about to say : This problem is trivial via the use of synthetic division.

139. anonymous

Blexting sec x is not 1 / sin x, btw.

140. anonymous

it's wrong blexting

141. anonymous

lol

142. anonymous

check mine please ^^" if I'm wrong, then I'm sorry

143. anonymous

My answer (which sstarica now has, too) has been right all along.

144. anonymous

ok this the last try i have left any final answers

145. anonymous

lol INew, you got the same answer? :)

146. anonymous

The answer =sec x + csc x - sin x =sec x + cos x cot x Now guess which is the required form :D Yes, got it above.

147. anonymous

yeah he did or she

148. anonymous

Oh, I'm a she. Hence the Newton name and everythi-- oh, wait a second ...

149. anonymous

lol, then loco go with it, I got the same answer too ^_^, damn my body temperature increased, too much concentration ^^"

150. anonymous

good question ._.

151. anonymous

nope it didn't work. sorry guys. told u the system is bs

152. anonymous

=-=......I'm sorry

153. anonymous

i got a 90 on the quiz though

154. anonymous

excellent :)

155. anonymous

Out of interest, what level is the quiz? As in, what educational level?

156. anonymous

THANKS FOR TRYING NEWTON (HE OR SHE?) AND SSTARICA (SEND ME UR # ;p)

157. anonymous

PRE-CALC

158. anonymous

159. anonymous

but the professor is really tough, doesn't curve or anything

160. anonymous

Oh, I'm english, so these american maths terms mean nothing to me ¬_¬ And trust me, that wasn't me trying.

161. anonymous

american ed system succks retricemate

162. anonymous

Don't worry, the person on the other thread got it wrong.

163. anonymous

bloody teachers

164. anonymous

yeah he did

165. anonymous

lol, relax, so what if the teaher sucks? depend on yourself and trust your abilities :), I had the worse teachers, so I had to depend on myself and look where I've reached :) good luck loco ^_^

166. anonymous

THANKS SSTARICA :)

167. anonymous

for my own personal interest, where have you reaced?

168. anonymous

np :)

169. anonymous

what do u mean

170. anonymous

"Look where I've reached" - I thought that meant a title/place, but now I think about I think it just means an ability level. My bad.

171. anonymous

i think it can mean both. depends on what you're talking about