# sinx+cosx=0

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• July 23rd 2011, 04:49 PM
OhCurtis
sinx+cosx=0
Solve for 0 <x< 360 degrees

Okay this is probably a stupid question but I'm really not doing so good with this unit so far.

Apparently to get the answer you have to take a squaring approach

sinx=-cosx
Square both sides
sin^2x=cox^2x
sin^2x=1-sin^2x
2sin^2x=1
sin^2x=1/2
sinx=+ or - (1/sqrt(2))
45, 135, 225, 315 degrees

BUT what I would have done is

sinx=-cosx
-sinx/cosx=cosx/cosx
-tanx=1
tanx=-1
135, 315 degrees

I need someone to tell me why my thinking process would or would not work. Thanks!

EDIT: Similar question I need help with too

sinx-cosx=1

Here's my approach
(sinx)^2=(1+cosx)^2
sin^2x=1+2cosx+cos^2x
1-cos^2x=1+2cosx+cos^2x
0=2cosx+2cos^2x
0=2cosx(1+cosx)
cosx=0,-1
90, 180, 270 degrees

However 270 is not an answer

I feel like I understand the concept just not getting the same answers. Any help would be GREATLY appreciated
• July 23rd 2011, 05:57 PM
pencil09
Re: sinx+cosx=0
Quote:

Originally Posted by OhCurtis
Solve for 0 <x< 360 degrees

Okay this is probably a stupid question but I'm really not doing so good with this unit so far.

Apparently to get the answer you have to take a squaring approach

sinx=-cosx
Square both sides
sin^2x=cox^2x
sin^2x=1-sin^2x
2sin^2x=1
sin^2x=1/2
sinx=+ or - (1/sqrt(2))
45, 135, 225, 315 degrees

BUT what I would have done is

sinx=-cosx
-sinx/cosx=cosx/cosx
-tanx=1
tanx=-1
135, 315 degrees

I need someone to tell me why my thinking process would or would not work. Thanks!

EDIT: Similar question I need help with too

sinx-cosx=1

Here's my approach
(sinx)^2=(1+cosx)^2
sin^2x=1+2cosx+cos^2x
1-cos^2x=1+2cosx+cos^2x
0=2cosx+2cos^2x
0=2cosx(1+cosx)
cosx=0,-1
90, 180, 270 degrees

However 270 is not an answer

I feel like I understand the concept just not getting the same answers. Any help would be GREATLY appreciated

hmm... remember the concept

Quadrant I ----> sin, cos, and tan (+)
Q II -----> sin (+)
Q III -----> tan (+)
Q IV -----> cos (+)

and try again... good luck...
• July 23rd 2011, 06:18 PM
skeeter
Re: sinx+cosx=0
Quote:

Originally Posted by OhCurtis
Solve for 0 <x< 360 degrees

Okay this is probably a stupid question but I'm really not doing so good with this unit so far.

Apparently to get the answer you have to take a squaring approach

sinx=-cosx
Square both sides
sin^2x=cox^2x
sin^2x=1-sin^2x
2sin^2x=1
sin^2x=1/2
sinx=+ or - (1/sqrt(2))
45, 135, 225, 315 degrees

BUT what I would have done is

sinx=-cosx
-sinx/cosx=cosx/cosx
-tanx=1
tanx=-1
135, 315 degrees

I need someone to tell me why my thinking process would or would not work. Thanks!

using tangent works fine if you remember that cosx cannot = 0

EDIT: Similar question I need help with too

sinx-cosx=1

Here's my approach
(sinx)^2=(1+cosx)^2
sin^2x=1+2cosx+cos^2x
1-cos^2x=1+2cosx+cos^2x
0=2cosx+2cos^2x
0=2cosx(1+cosx)
cosx=0,-1
90, 180, 270 degrees

However 270 is not an answer

solving an equation by squaring both sides sometimes introduces extraneous (invalid) solutions ... that's why you need to check all solutions when you use this technique. the steps you took were fine ... just remember to check the solutions.

I feel like I understand the concept just not getting the same answers. Any help would be GREATLY appreciated

...
• July 24th 2011, 12:55 AM
Siron
Re: sinx+cosx=0
In general you can say for trigoniometric equations:
$\sin(x)=\sin(A)$
Two solutions:
- $x=A + 2k\pi$
- $x=\pi - A + 2k\pi$

$\cos(x)=\cos(A)$
Two solutions:
- $x=A + 2k\pi$
- $x=-A+2k\pi$

$\tan(x)=\tan(A)$
one solution
- $x=A+k\pi$

Where $k \in \mathbb{Z}$
• July 24th 2011, 02:16 AM
Archie Meade
Re: sinx+cosx=0
As skeeter stated, after squaring both sides you must check the solutions.
For example, in the first quadrant, both Sinx and Cosx are non-negative
and hence they cannot sum to zero.
Cosx is zero at multiples of 90 degrees and Sinx is zero at 0 degrees and multiples of 180 degrees,
so they are never zero for the same x.

Solutions to Cosx+Sinx=0 require they be opposite signs,
which occurs in quadrants 2 and 4.
Also, their numerical magnitudes (moduli) must be equal in order to cancel and get zero,
and hence they lie at the point of intersection of the line of slope -1 going through the origin
and the circumference of the unit circle.
• July 24th 2011, 07:21 AM
OhCurtis
Re: sinx+cosx=0
All of that explained it so much better than my book. Thanks for all the help!