Question

sinx+sin3x=cosx

Answer 1

sin(3x) = sin(2x + x)

Answer 2

the value of sin3x is 3sinx-4sin^3x

Answer 3

ok lets substitute that

Answer 4

I'm stuck with 4sinx-4sin^3 x = cosx

Answer 5

thanks for the reply! btw

Answer 6

yw

Answer 7

so, what happens next?

Answer 8

unfortunately we dont have an expression in terms of cos x alone (or sin x alone)

Answer 9

so there is no answer?

Answer 10

you can graph y = 4sinx-4sin^3 x, y = cosx and find the intersection points

or you can graph

you can graph 4sinx-4sin^3 x - cosx = 0 , and find the roots

Answer 11

hmmm, what if we use the sum and difference of cosines? would there be an answer?

Answer 12

someone pls answer

Answer 13

let me try wolfram, sometimes they have a good idea to use

Answer 14

bear with me, slow computer

Answer 15

its, ok ; just thought you left me a while ago hahaha :)

Answer 16

now my computer is frozen :)

Answer 17

sure

Answer 18

i might have to do a hard reboot.
mathematica uses something online to fetch information

Answer 19

yeah, its fine no need to rush :)

Answer 20

ok did a hard reboot

Answer 21

yes there is a way to simplify this

Answer 22

its amazing they can program this stuff

Answer 23

Solve for x:
sin(x)+sin(3 x)==cos(x)


Subtract cos(x) from both sides:
-cos(x)+sin(x)+sin(3 x)==0


Simplify trigonometric functions:
cos(x) (2 sin(2 x)-1)==0


Split into two equations:
cos(x)==0 or 2 sin(2 x)-1==0

Answer 24

What's with the double ==? Haha.

Answer 25

i dont know

Answer 26

wolfram syntax i guess

Answer 27

Interesting.

Answer 28

what program did you use?

Answer 29

this is mathematica 9

Answer 30

woah thanks

Answer 31

The question seems like you are trying to prove one side equals the other.

Answer 32

actually its not an identity, so then you must be looking for solutions

Answer 33

sin(2) + sin(6) =/= cos(2)

Answer 34

the directions were find all the solutions in the interval of 0<=x<=2pi

Answer 35

so let me supply the missing steps here

Answer 36

ok, thanks again!!!

Answer 37

Oh, then the whole question was not even up there lol

Answer 38

ok its obvious that dr. wolfram went from

sin x + sin 3x - cos x = 0
to
cos x (2 sin 2x - 1 ) = 0

we want to supply the missing steps

Answer 39

how?

Answer 40

its not obvious how wolfram got that

Answer 41

i meant, its obvious that wolfram did something :)

Answer 42

he did some kind of expansion

Answer 43

yeah , any idea how?

Answer 44

we can try expanding this

sin x + sin 3x - cos x = 0

sin x + sin (2x + x ) - cos x = 0

sin x + (sin2x cosx + sin x cos 2x) - cos x = 0

Answer 45

sin x + 2 sin x cos x cos x + sin x ( 2 cos^2(x) - 1 ) - cos x = 0

Answer 46

yeah but to get
cos x (2 sin 2x - 1 ) = 0
from
sin x + sin 3x - cos x = 0
we need to have 2sin2xcosx-cosx right?

Answer 47

does it make sense so far, these steps

sin x + sin 3x - cos x = 0

sin x + sin (2x + x ) - cos x = 0

sin x + (sin2x cosx + sin x cos 2x) - cos x = 0

sin x + 2 sin x cos x cos x + sin x ( 2 cos^2 x - 1 ) - cos x = 0

Answer 48

yeah, I'm still not lost surprisingly lol

Answer 49

i used cos (2x) = 2 cos^2 ( x ) - 1 , substitution

Answer 50

yeah i see that

Answer 51

That makes a lot more sense.

Answer 52

so it would be 4 sin x cos^2(x) -cos x?

Answer 53

sin x + sin 3x - cos x = 0

sin x + sin (2x + x ) - cos x = 0

sin x + sin2x cosx + sin x cos 2x - cos x = 0

sin x + 2 sin x cos x cos x + sin x ( 2 cos^2 x - 1 ) - cos x = 0

sin x + 2 sin x cos^2(x) + 2sinx cos^2(x) - sin x - cos x = 0

4 sin x cos^2(x) - cos x = 0

Answer 54

hmmm, there is a 4

Answer 55

oh maybe wolfram got 4sinx(cosx*cox) -cosx

Answer 56

ok i see now

Answer 57

then since 2 cosx sin = sin2x
4sinx(cosx*cox) -cosx
2(2sinxcosx)cosx-cosx
2sin2x cosx-cosx?

Answer 58

sin x + sin 3x - cos x = 0

sin x + sin (2x + x ) - cos x = 0

sin x + sin2x cosx + sin x cos 2x - cos x = 0

sin x + 2 sin x cos x cos x + sin x ( 2 cos^2 x - 1 ) - cos x = 0

sin x + 2 sin x cos^2(x) + 2sinx cos^2(x) - sin x - cos x = 0

4 sin x cos^2(x) - cos x = 0

2 *cos x * 2 sin x cos x - cos x = 0

2cos x * sin(2x) - cos x = 0


cos x ( 2 sin(2x) - 1) = 0

Answer 59

yeahhhhhh! so its right?

Answer 60

it would be nice if we could condense this argument, but yes it works

Answer 61

seems like a lot of steps :/

Answer 62

and we still have to set them equal to zero

Answer 63

hmm, last question...
how do you get the value of 2 sin(2x) - 1?

Answer 64

but we can have the two answers right?
cosx l ( 2 sin(2x) - 1)=0
x=cos^-1 (0) l

Answer 65

right

Answer 66

2 sin(2x) - 1 = 0

2 sin (2x) = 1

sin(2x) = 1/2

2x = sin^-1 (1/2)

x = 1/2 * sin^-1 (1/2)

Answer 67

so it will be sin^-1 (1/4)?

Answer 68

you can't multiply 1/2 and 1/2 there

Answer 69

oh ok

Answer 70

x = 1/2 * sin^-1 (1/2)
x = 1/2 * { pi/6 +2pi * n , 5p/6 + 2pi*n}

Answer 71

so in degree form, the anser is 195 and 255?

Answer 72

x = 1/2 * sin^-1 (1/2)
x = 1/2 * { pi/6 +2pi * n , 5p/6 + 2pi*n}

carefully distribute the 1/2

x = 1/2 * pi/6 +2pi/2* n , 5p/6*1/2 + 2pi/2*n

x = pi/12 +pi* n , 5p/12 + pi*n

Answer 73

I just realized there is a simpler way to solve this, than the expansion above

Answer 74

there is a formula called
sum to product formula

Answer 75

how? but I'm already satisfied with this but can i see that formula?

Answer 76

sure

Answer 77

http://www.sosmath.com/trig/Trig5/trig5/trig5.html

Answer 78

scroll down to sum to product formula

Answer 79

yeah, I already know that formula but I still can't see the application?

Answer 80

sinx+sin3x=cosx

we can use it on the left side

sin ( 3x + x) = 2 sin ((3x + x )/2)* cos ((3x - x )/2)

Answer 81

okay

Answer 82

ok

Answer 83

sin ( 3x + x)
= 2 sin ((3x + x )/2)* cos ((3x - x )/2)
= 2 sin ((4x) /2) cos((2x)/2)

Answer 84

then?

Answer 85

sinx+sin3x=cosx

sin x + sin ( 3x + x)= cos x

sin x + 2 sin ((3x + x )/2)* cos ((3x - x )/2) = cos x

sin x + 2 sin (2x) cos(x) = cos x

Answer 86

ok ok

Answer 87

sorry i thought it would have made it easy to solve :)

Answer 88

if i see a simpler approach i will post it here

Answer 89

are you ok with the solutions

Answer 90

yeah, and btw the long solution is fine with me just looking for other ways to solve it :) and THANKS A LOT!!!

Answer 91

oh

Answer 92

dan did i miss something insightful (aka easier approach)

Answer 93

STOLEN!

Answer 94

so i have to change the product to a sum

Answer 95

oh sry i think i reverse identity

Answer 96

1/2(sin(2x+x)+sin(2x-x))=sin(2x)cos(x)
(sin(2x-x)+sin(2x+x))=2sin(2x)cos(x)=cos(x)
sin(2x)=1/2
2x=pi/6 or 5pi/6
x=pi/12 +2pin or 5pi/6+2pin, n E Z

Answer 97

nvm this is fine, is that what u got aswell?

Answer 98

were you continuig where i left off?