# P2 Pure maths - TrigWatch

This discussion is closed.
Thread starter 13 years ago
#1
sin(3x) = cos (x)
find between 0 and two pi

Please can you show steps, I cant seem to establish the correct method
0
13 years ago
#2
(Original post by w0rldl3ad3r)
sin(3x) = cos (x)
find between 0 and two pi

Please can you show steps, I cant seem to establish the correct method
There are several methods

(1) By inspection I see that Ā¼(pi) is a solution.

(2) Draw the graphs sin(3x) and cos(x) and see where they intersect.

(3) By an iteration formula; xn+1 = cos-1(sin(3xn)).

(4) By algebraic manipulation.
0
13 years ago
#3
(Original post by w0rldl3ad3r)
sin(3x) = cos (x)
find between 0 and two pi

Please can you show steps, I cant seem to establish the correct method
You can use the identity > sinĀ²x + cosĀ²x = 1

so sinx = √(1- cosĀ²x)

sub sinx = √(1- cosĀ²x):

√(1- cosĀ²x) - cosx = 0

Take the square of both sides to get:
1-2cosĀ²x = 0

Now just rearrange to solve for x:

cosx = (√(1/2))
x = arccos(√(1/2))
x = pi/4, pi/8 and 5pi/4
0
13 years ago
#4
(Original post by w0rldl3ad3r)
sin(3x) = cos (x)
find between 0 and two pi

Please can you show steps, I cant seem to establish the correct method
I couldn't solve it. But this is what I tried.

Bloody hell. Erm let me see...

sin(3x) = sin(2x + x) = sin2xcosx + sinxcos2x

and sin(2x) = 2sinxcosx

and cos(2x) = 2cosĀ²x - 1 = 1 - 2sinĀ²x

=> sin(3x) = 2sinxcosĀ²x + sinx(2cosĀ²x - 1)

=> 2sinxcosĀ²x + 2sinxcosĀ²x - sinx => 4sinxcosĀ²x - sinx

Now put 4sinxcosĀ²x - sinx = cosx

But: cosĀ²x = 1 - sinĀ²x

=> 4sinx(1 - sinĀ²x) - sinx = cosx

=> 3sinx - 4sinĀ³x = cosx

=> 3tanx - 4tanxsinĀ²x = 1

=> NO IDEA!

Sorry.
0
13 years ago
#5
(Original post by Vijay1)
You can use the identity > sinĀ²x + cosĀ²x = 1

so sinx = √(1- cosĀ²x)

sub sinx = √(1- cosĀ²x):

√(1- cosĀ²x) - cosx = 0

Take the square of both sides to get:
1-2cosĀ²x = 0

Now just rearrange to solve for x:

cosx = (√(1/2))
x = arccos(√(1/2))
x = pi/4 and 5pi/4
Where is the 3x?
0
13 years ago
#6
(Original post by Vijay1)
You can use the identity > sinĀ²x + cosĀ²x = 1

so sinx = √(1- cosĀ²x)

sub sinx = √(1- cosĀ²x):

√(1- cosĀ²x) - cosx = 0

Take the square of both sides to get:
1-2cosĀ²x = 0

Now just rearrange to solve for x:

cosx = (√(1/2))
x = arccos(√(1/2))
x = pi/4 and 5pi/4
It asks for sin(3x), not sin(x).

If you were given

sin(x) = cos(x) then you wouldn't have to go through all the steps above, just note that

sin(x) = cos(x)
tan(x) = 1

x = arctan(1)
x = Ā¼(pi), (5/4)(pi), ...
0
13 years ago
#7
sin(3x) = 4.sin(x).cos2(x) - sin(x)

=> 4.√(1 - cos2(x)).cos2(x) - √(1 - cos2(x)) = cos(x)

I know this looks rather hidious, but solving this will give you

x = Ā¼(pi), (1/8)(pi), (5/8)(pi)

________________

Alternatively,

sin2(3x) + cos2(3x) = 1

sin(3x) = √(1 - cos2(3x))

So

1 - cos2(3x) = cos2(x)

and take note of the fact that

cos2(3x) = (4.cos3(x) - 3.cos(x))2

and then solve for cos(x).
0
13 years ago
#8
(Original post by samdavyson)
Where is the 3x?
Because the statement is that sin3x = cosx

we must understand that the value of sine of 3x will be the same as the value of cosine of x. Hence the identity of

sinĀ²x + cosĀ²x = 1

still applies, regardless of the constant 3.
0
13 years ago
#9
Erm.. Yes, the identity sinĀ²x+cosĀ²x=1 always holds. But that doesn't mean:
sin3x = cosx => cosx = sinx
0
13 years ago
#10
the idetity cosĀ²x + sinĀ²x = 1 is always true but ONLY when the coef. of x are the same.
i.e. cosĀ²x + sinĀ²x = 1
cosĀ²3x + sinĀ²3x = 1
cosĀ²100x + sinĀ²100x = 1
cosĀ²nx + sinĀ²nx = 1
0
13 years ago
#11
I know what you all mean but what I mean is this:

sin3x, after solving it we find pi/4 >> sin3(pi/4) = 0.707...

cosx, after solving it we find pi/4 >>> cos(pi/4) = 0.707...

hence cosx = sin3x

so we can use the identity
0
13 years ago
#12
(Original post by Vijay1)
Because the statement is that sin3x = cosx

we must understand that the value of sine of 3x will be the same as the value of cosine of x. Hence the identity of

sinĀ²x + cosĀ²x = 1

still applies, regardless of the constant 3.
The function sin(x) differs to the function sin(3x).

It is true that sinĀ²x + cosĀ²x = 1, but what has this got to do with the implication that sin(x) = cos(x)? You cannot deduce this from sinĀ²x + cosĀ²x = 1.
0
13 years ago
#13
sin3x, after solving it we find pi/4 >> sin3(pi/4) = 0.707...

cosx, after solving it we find pi/4 >>> cos(pi/4) = 0.707...
how would u solve them to find pi/4?
0
13 years ago
#14
(Original post by Vijay1)
I know what you all mean but what I mean is this:

sin3x, after solving it we find pi/4 >> sin3(pi/4) = 0.707...

cosx, after solving it we find pi/4 >>> cos(pi/4) = 0.707...

hence cosx = sin3x

so we can use the identity
Your implication is the wrong way round.

You cannot say

sin3(pi/4) = 0.707... cos(pi/4) = 0.707...

=> cosx = sin3x.

But you can say

cosx = sin3x

=> x = (pi/4).
0
13 years ago
#15
OK guys, I know what you all mean!
My sines have led to the wrong cosines.

It was due to the nature of the question and shear luck that I got the answer correct, and ignore what I said in my last post.
The truth is I forgot about substituting the 3 back into the equation, but didn't bother to correct it because the answer was right.

Your right, I'm wrong.
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