Question

Use an addition or subtraction formula to find the exact value of the expression.

sin(285)

Answer 1

Well, lets assume you know the 'obvious' values of sin - {0, 30, 45, 60, 90} + 360n.

How can you use them to make 285?

Answer 2

add them?

Answer 3

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.

Answer 4

so what do i add?

Answer 5

That's where i'm cofused

Answer 6

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

Answer 7

¬_¬
Sin 150 = Sin 30
Sin 135 = Sin 45

Answer 8

I do not get it.

Answer 9

Do you not get why that is true?

Answer 10

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

Answer 11

both

Answer 12

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.

Answer 13

ok thanks

Answer 14

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.

Answer 15

-90?

Answer 16

Last line is -sin(75).

Where did you get 90 from?

Answer 17

i have no idea what -sin 35= ?

Answer 18

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)

Answer 19

the answer is?

Answer 20

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

Answer 21

sin(45) = cos(45) = 1/sqrt(2)
sin(30) = 1/2
cos(30) = sqrt(3)/2

\[\implies -\sin(75) = -\frac{1}{\sqrt{2}} \cdot \left(\frac{1}{2} + \frac{\sqrt{3}}{2}\right) = -\frac{1+ \sqrt{3}}{2\sqrt{2}} = -\frac{\sqrt{2} + \sqrt{6}}{4} = \sin(285)\]

Answer 22

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

Answer 23

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 :)

Answer 24

any *

Answer 25

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

Answer 26

lol, own who?

Answer 27

the graph is actually easier, never realized it :p

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

thanks guys :)

Answer 34

np ^_^

Answer 35

but did you figure it out ? :)

Answer 36

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.

Answer 37

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 ^_^

Answer 38

stop fighting guys

Answer 39

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

Answer 40

Simplify the trigonometric expression.

cos[x]/sec[x] + sin[x]/csc[x]

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

how about we i mean u guys solve this problem

Answer 45

you still didn't solve it? .-.

Answer 46

no, i hate this stuff

Answer 47

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

Answer 48

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

Answer 49

\[\frac{\cos x}{\sec x} = \frac{\cos x}{\frac{1}{\cos x}} = \cos^2x\]

Answer 50

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

Answer 51

sstarica help?

Answer 52

hold on :)

Answer 53

newton it's wrong

Answer 54

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

Answer 55

sorry, i thought that was the answer

Answer 56

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

Answer 57

try to explain it in a simple way :)

Answer 58

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.

Answer 59

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

Answer 60

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

Answer 61

he posted another question?

Answer 62

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

Answer 63

Thankfully still no graphs required

Answer 64

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

Answer 65

u can write it here

Answer 66

nvm it, INew answered it though :)

Answer 67

it's not right though

Answer 68

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 ^_^

Answer 69

you are right :) i got the same answer

Answer 70

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

Answer 71

well done ^_^

Answer 72

lol, I didn't get you INew

Answer 73

\[\frac{1}{2} \geq 1/2\]

Answer 74

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

Answer 75

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

Answer 76

last problem....

Simplify the trigonometric expression.

cos[x]/ tan[x] + sex[x]

Answer 77

frac{a}{b} gives\[\frac{a}{b}\]

Answer 78

oh, thank you :)

Answer 79

sex(x) - lolololololol

Answer 80

LOL

Answer 81

sec[x], sorry :P

Answer 82

wish math was that fun

Answer 83

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

Answer 84

When trying to simplify things.

Answer 85

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

^_^

Answer 86

ORLY?

Answer 87

maybe? :) correct me lol

Answer 88

it's not correct

Answer 89

\[\cos^2x + 1 \not= \sin^2x\]

Answer 90

hmm, where is my mistake

Answer 91

LOL! right >_< sorry

Answer 92

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

Answer 93

hmm, solve it :)

Answer 94

man, u guys are at it.

Answer 95

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

Answer 96

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