How do you convert sin into cos, and vice versa.
Promblem: cos(6x) = sin(3x-9)

- anonymous

How do you convert sin into cos, and vice versa.
Promblem: cos(6x) = sin(3x-9)

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- anonymous

cos(6x)= sin(90-6x)= sin(3x-9)
sin values are equal u can equate the angles
90-6x=3x-9
99=9x
x=11

- anonymous

i don't think this that easy
take a look at the solutions
http://www.wolframalpha.com/input/?i=cos%286x%29%3Dsin%283x-9%29

- anonymous

I'm not sure if it's with radians or not but I'm in trig if that helps any. I'm just looking for a conversation rate if there is one.

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## More answers

- Hero

6x + 3x - 9 = 90

- ZeHanz

You want to convert to an equation of the form cos a = cos b or sin a = sin b, because these kind of equations have standard solutions:\[\cos a=\cos b \Leftrightarrow a= \pm b+2k \pi\]\[\sin a = \sin b \Leftrightarrow a=b+2k \pi \vee a=\pi-b+2k \pi\](assuming radians)
To get to one of these forms, you can use the following conversions:\[\cos x=\sin(\frac{ \pi }{ 2 }-x)\]or \[\sin x =\cos(\frac{ \pi }{ 2 }-x)\]

- Hero

x = 11
sin(66) = sin(24) in degrees

- anonymous

@ZeHanz So you're saying I would take cos(6x) and subtract 6x from pi/2 giving me sin(pi/2 - 6x) being equal to cos(6x)

- anonymous

@Hero I appreciate the answer but I'm looking for an explanation

- Hero

sin(x) = cos(y) are co-functions, meaning angles x and y are complementary.

- ZeHanz

@RobZBHayes : exactly!

- anonymous

@ZeHanz alright so sin(pi/2 - 6x) = sin(3x-9) and cos(6x) = sin(3x-9) are the same problem, correct? So how would I go about solving sin(pi/2 - 6x) = sin(3x-9)?

- Hero

You mean it is the same type of problem, right?

- Hero

But not the same exact problem.

- Hero

For sin(pi/2 - 6x) = sin(3x - 9) the angles are equal so
pi/2 - 6x = 3x - 9

- Hero

Solve for x

- anonymous

I tried that and came up with an answer that isn't in my answer choices. I got 0.67453292519943295769236907684886, I'm not sure if it's correct or not though. First I added 6x to both sides giving me pi/2=9x-2, then I multiplied everything by 2 giving me pi=18x-9, after that I added 9 to both sides leaving me with 12.14159265358979323846264338328=18x, finally I divided everything by 18 giving me x=0.67453292519943295769236907684886. Did I do all of that correctly? @Hero @ZeHanz

- Hero

What are your answer choices?

- anonymous

x=10
x=-10
x=11
x=12
You said before that the answer is 11 and I think that's right but I'm trying to get the relation between sin, cos, and tan in case I come across another problem like this one.

- Hero

Yes, that's what I got

- Hero

pi/2 - 6x = 3x - 9
180/2 - 6x = 3x - 9
90 + 9 = 6x + 3x
99 = 9x
99/9 = x
11 = x

- Hero

Remember, pi = 180

- Hero

I already explained to you the relationship
sin(x) = cos(y) if x and y are complementary
tan(x) = cot(y) if x and y are complementary
csc(x) = sec(y) if x and y are complementary

- Hero

Those are properties of co-functions

- anonymous

So the covertion of sin to cos and vice versa is always pi/2-x and pi always equals 180?

- Hero

pi = 180 because the length of pi is a semicircle having a measure of 180 degrees.

- anonymous

Oh ok, I wasn't aware that pi was a semicircle lol. I guess that's kind of sad considering I'm in trig -_-

- ZeHanz

I guess "being in trig" means you have to do everything in degrees and do not need to worry about angles greater than 180 degrees (or 360?) and smaller than 0?
In that case I would solve the equation as follows:\[\cos 6x=\sin (3x-9) \Leftrightarrow \cos 6x=\cos(90 - (3x-9))=\cos((99-3x)\]This would give:\[6x=\pm (99-3x)\]which means\[6x=99-3x \vee 6x=3x-99 \Leftrightarrow\]\[9x=99 \vee 3x=-99 \Leftrightarrow x=11 \vee x=-33\]So x=11 would be the only answer that is left, so option C

- anonymous

So if I want to solve cos(2x-4)=sin(6x) the first thing I would want to do is convert sin(6x) into a cos, and that would be cos(pi/2-6x). Then my equation would be 2x-4=pi/2-6x, so if I were to simplify that It would be 2x-4=90-6x, then I would want to subtract 2x from both sides giving me -4=90-8x, then I would subtract 90 from both sides giving me -94=-8x, finally I would want to divide everything by -8x giving me my final answer as x=11.75 as a final answer, right?

- ZeHanz

Yes, but there is no need to use radians first and then degrees. If you do this, you make things much more difficult for yourself.Either do everything in degrees, or do evertything in radians.
Further, I'm still unsure about what "being in trig" means.
I would get:\[2x-4= \pm 90-6x\]so\[2x-4=90-6x \vee 2x-4=6x-90 \Leftrightarrow\]\[8x=94 \vee 4x=86 \Leftrightarrow\]\[x=11.75 \vee x=21.5\]The last solution coming from the negative solution, which can probably be discarded...

- anonymous

Trig is a shortened way of saying Trigonometry

- ZeHanz

Which means every angle is an angle in a triangle?

- Hero

If you wanted to solve cos(2x-4)=sin(6x)
knowing that angles 2x - 4 and 6x are complementary you would simply add both equations together and set them equal to 90
2x - 4 + 6x = 90
8x - 4 = 90
8x = 94
x = 94/8
x = 11.75 in degrees.

- Hero

@ZeHanz, your approach seems to include unnecessary steps.

- Hero

All one needs to know is that if sin(x) = cos(y) then you are dealing with co-functions where x and y are complementary.

- anonymous

Our of curiosity how would you determine if the angles are in fact complementary in these types of equations?

- Hero

If you see a situation where you have an equation of the form
sin(x) = cos(y), then x and y must be complementary.

- Hero

The equation in that form is the definition of a co-function. That's how you know.

- Hero

It's not the mystery you're making it out to be. Co-functions are simply a property of trigonometric functions

- Hero

http://www.barstow.edu/lrc/tutorserv/handouts/045%20Trigonometric%20Properties.pdf

- Hero

sin(x) = cos(y) is a general form of a co-function identity where x and y are complementary angles such that x + y = 90
sin(6x) = cos(2x - 4) is a specific form of the co-function identity where solving 6x + (2x - 4) = 90 for x reveals the complementary angles.

- Hero

It's as simple as that.

- ZeHanz

@Hero: I understand now these equations are not about the sin and cos functions defined for every real number, but that they are connected to sin and cos as defined in triangles.
That explains my "unnecessary steps" ;)

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