• Want to help us with this year's BoS Trials?
    Let us know before 30 June. See this thread for details
  • Looking for HSC notes and resources?
    Check out our Notes & Resources page

Trig question: transformation (1 Viewer)

mrman

Member
Joined
Apr 21, 2003
Messages
98
can someone find the exact value of

sinx-cosx=1

-pi<(or equal to)x<(or equal to)pi


The problem is I dont exactly know what to do, with negative pi, and what valuews of x u get with that.

I know im pretty screwed half yearly in like 5 days lol!!!!!


NOOOOOOOOOOO:D :D :D :D
 

kini mini

Active Member
Joined
Sep 19, 2002
Messages
1,272
Location
Sydney
Gender
Undisclosed
HSC
2002
Originally posted by mrman

The problem is I dont exactly know what to do, with negative pi, and what valuews of x u get with that.
In that case, I'll give you clues rather than tell you what to do...

A clue...that indicates the domain in which you have to find solutions :).

Another clue: sketch the y= sin x and y = cos x graphs on the same axes for the given domain, and see if you understand the question any better :)

Then see if you can solve algebraically :).
 

chunder

Member
Joined
Apr 9, 2003
Messages
131
Gender
Male
HSC
2003
what you do is express sinx-cox=1 in the form of rsin(x-a)=1

i.e. let sinx-cosx=rsinxcosa-rcosxsina=1

by equating: rcosa=1 and rsina=1

therefore r = root 2
in addition since tan a=1, a=45

therefore sinx-cosx=(root 2)sin(x-45)=1

etc etc etc

your answers will be 90 and 180
 

wogboy

Terminator
Joined
Sep 2, 2002
Messages
653
Location
Sydney
Gender
Male
HSC
2002
Alternatively, you can square both sides to get the solutions.

sinx - cosx = 1

sin^2x - 2sinxcosx + cos^2x = 1

1 - 2sinxcosx = 1

2sinxcosx = 0

sinx = 0 or cosx = 0 (or if you like, you can say sin2x = 0)

so the possibilities for x are:

x = pi
x = 0
x = -pi
x = pi/2
x = -pi/2

now because we squared both sides of the equation initially, you have to verify each answer (otherwise we can prove -1 = 1 :) )

sin(pi) - cos(pi) = 1, so x = pi is true.
sin(0) - cos(0) = -1, so x = 0 is false.
sin(-pi) - cos(-pi) = 1, so x = -pi is true.
sin(pi/2) - cos(pi/2) = 1, so x = pi/2 is true.
sin(-pi/2) - cos(-pi/2) = -1, so x = -pi is false.

so the final solutions are,

x = pi, x = -pi, x = pi/2
 
Last edited:

mrman

Member
Joined
Apr 21, 2003
Messages
98
Now i get it

Thanks heaps wogboy, Now I get it. Lol kini mini, I know what you do with the domain, but it was a bit hazy because in the worked solutions they said -5pi/4 and Im like perhaps im retarded and dont know what a domain is lol. But thanks all of you guys for helping....


much obliged ;):D :D
 

kini mini

Active Member
Joined
Sep 19, 2002
Messages
1,272
Location
Sydney
Gender
Undisclosed
HSC
2002
Originally posted by chunder
your answers will be 90 and 180
If the question is framed in radians, then you must answer in radians. What about the negative solutions? I did like wogboy, except I shifted cos x to the RHS and had it all in terms of cos :).

mrman , where does -5pi/4 come from :confused: ?
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top