Question

Find the exact values of sin(2θ), cos(2θ), and tan(2θ) for the given value of θ.

sec(θ)=-3/2 ; 90<θ<180

sin(2θ) =
cos(2θ) =
tan(2θ) =

how the heck am i gonna solve this..?

Answer 1

find theta with sec(theta) = -3/2

Answer 2

huh.. i didnt get it..?

Answer 3

do you know that sec is 1/cos?

Answer 4

reciprocal identity..

Answer 5

so it will be -2/3..?

Answer 6

yes cos = that

Answer 7

so i will use double angle identity..?

Answer 8

what? no

Answer 9

i dont really get it.. am i just gonna plug it in or what..?

Answer 10

oh i see you meant for the sin(2theta) yes you will use double angle

Answer 11

also are you sure the question doesn't say sec(theta) = -sqrt(3)/2?

Answer 12

sigh* i thought i was wrong.. hahah.. ok.. double angle just plug it or need one more to solve the exact value..?

Answer 13

nope

Answer 14

so the cos(2theta) = -1/9..?

Answer 15

yes

Answer 16

how am i suppose to solve tan(2theta) where it is equal to 2tan(theta)/1-tan(theta)..?

Answer 17

or because tan is equal to sin/cos

Answer 18

lets start with sin

Answer 19

sin(arccos(x) = sqrt(1-x^2)

Answer 20

ow yeah.. the sin.. hahah.. forgot about that.. how am i gonna solve for sin it is equal to sin(2theta)=2sin(theta)cos(theta).. sin is missing and sin is needed to solve sin..? mind blown..

Answer 21

whoa whoa whoa what is arccos..?

Answer 22

\[\cos^{-1} \]

Answer 23

ahhh.. ok hahha.. thats arcos.. raised to -1

Answer 24

so knowing that \[\sin (\cos^{-1} (x) = \sqrt(1-x^2)\]

Answer 25

yes..

Answer 26

to solve x i need to get x y r..?

Answer 27

\[\sin (2 \cos^{-1} (-2/3))=\sqrt(1-(-2/3)^2)*\cos(\cos^{-1} (-2/3))\]

Answer 28

now it gets complicated..

Answer 29

no no btw slow computer so sorry but theta = cos^-1(-2/3) right?

Answer 30

yep

Answer 31

okay so \[\sin (2\theta)=2\sin (\theta)\cos(\theta)\]

Answer 32

so it is ewual to -2 square root of 5/9

Answer 33

substitute cos^-1(-2/3) with theta

Answer 34

i mean \[-\frac{ 2\sqrt{5} }{ 9 }\]

Answer 35

almost i forgot the 2 in 2sinxcosx!

Answer 36

so u mean theta is equal to cos^-1(-2/3)..?

Answer 37

so your answer times two! and you are great!

Answer 38

yes

Answer 39

how did you solve the cos(2x) then?

Answer 40

this one..? cos(2theta) = -1/9..

Answer 41

yes how did you solve it without cos^-1

Answer 42

coz cos(2theta) is also equal to 2cos(theta)^2-1

Answer 43

XD oh you don't need cos^-1 for that....

Answer 44

okay so you have sin and cos now for tangent!

Answer 45

solve sin/cos!

Answer 46

weyt u were way to fast.. were are we now..? haha

Answer 47

\[\frac{ \sqrt(1-(-\frac{ 2 }{ 3 }) }{ -2/3 }\]

Answer 48

oh im sorry, remember we did sin it was that crazy number \[\frac{ -4\sqrt(5) }{ 9 }\]

Answer 49

\[-\frac{ \sqrt{15} }{ 2 }\] so is this it..?

Answer 50

not 15 just 5

Answer 51

my calculator says its 15..

Answer 52

....i keep screwing up the formula

Answer 53

what..? hahah.. and by the way its -2 not -4.. haha

Answer 54

\[-\frac{ 2\sqrt{5} }{ 9 }\] thats it not -4

Answer 55

again i screwed up the formula that time too i told you there was an extra TIMES 2 that i forgot

Answer 56

hahaha.. now im confused.. hahaha

Answer 57

so u mean i need to multiply it to 2 or not..?

Answer 58

take your \[\frac{ -2\sqrt5 }{ 9 }\] and multiply it by 2

Answer 59

you only solved sin(2x) = sin(x)cos(x) but it should be 2 sin(x)cos(x)

Answer 60

nevermind just trust me

Answer 61

so u said that u forgot to say that i need it to multiply to 2..?
hHAHAHAH.. i know i trust u.. hahaha ok where are we.. on the -4 which is it is the sin..?

Answer 62

if you want reinput these formulas
sin(2x)=\[2\sqrt{1-\frac{ -2 }{ 3 }}*-\frac{ 2 }{ 3 }\]

Answer 63

ok so that is sin..?

Answer 64

and u are right it is -4

Answer 65

here is where we are
\[\sin(2x)= \frac{ -4\sqrt5 }{ 9 }\]
\[\cos(2x) = -\frac{ 1 }{ 9 }\]

Answer 66

how about the \[-\frac{ \sqrt{15} }{ 9 }\] is it 15 or 5..?

Answer 67

okay let me put in the correct one this time

Answer 68

\[\frac{ \sqrt{1-(\frac{ -2 }{ 3 })^2} }{ 2/3 }\]

Answer 69

equals \[-\frac{ \sqrt5 }{ 2 }\]

Answer 70

this is TAN(X) not TAN(2X) so all you need to do is what?

Answer 71

\[\frac{ 2\tan(x) }{ 1-\tan(x)^2 }\]

Answer 72

x is -sqrt 5/2..?

Answer 73

or solve tan then plug it in that equation..?

Answer 74

yes yes but guess what!!

Answer 75

what..?

Answer 76

we just now did that!

Answer 77

whoa..? really how..?

Answer 78

tan is \[-\sqrt5/2\]

Answer 79

hwahaha.. that was unexpected.. hahaha..

Answer 80

that crazy formula was sin/cos

Answer 81

i just reformed it

Answer 82

hey can i ask a favor..?

Answer 83

sure

Answer 84

from what country are u..?

Answer 85

merica

Answer 86

why

Answer 87

im from philippines.. can u be online tomorrow..? what time is it in your america..?

Answer 88

2:25AM

Answer 89

crazy that im still up huh?

Answer 90

oh um.. can u be online earlier like 12 am..? gamer peeps..?

Answer 91

12 am my time?

Answer 92

yep is that ok..? in our time its like 4 pm..

Answer 93

yea sure XD

Answer 94

yey.. coz i need to log out now.. see u tomorrow i need help about that again tomorrow.. by the way.. thanks.. im expecting u tom.. thanks again.. :)

Answer 95

bye

Answer 96

bye