Trigonometric Identities algebra

  • Thread starter Mspike6
  • Start date
  • #1
63
0
hello.

i have a question about trignometric identities.. it's realtivly easy, but am struggling with the algebra in it ( Algebra + trig = Very confusing to me )


Prove that ..

[Sinx/(1+Cosx)] + [(1+cosx) / sinx] = 2csc x



i manged to get it to [Sin2x+1+2cosx+cos2x] / SinxCosx

but i don't know what to do after that

any help would be appreciated, even if hints or steps without actually solving it for me .

Thank you

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
berkeman
Mentor
60,564
10,883
hello.

i have a question about trignometric identities.. it's realtivly easy, but am struggling with the algebra in it ( Algebra + trig = Very confusing to me )


Prove that ..

[Sinx/(1+Cosx)] + [(1+cosx) / sinx] = 2csc x



i manged to get it to [Sin2x+1+2cosx+cos2x] / SinxCosx

but i don't know what to do after that

any help would be appreciated, even if hints or steps without actually solving it for me .

Thank you

Homework Statement





Homework Equations





The Attempt at a Solution


There's an identity involving sin2x and cos2x that might help get you a bit farther...
 
  • #3
867
0
For the first term sinx/(1 + cosx), multiply everything by the conjugate of the denominator. It should be a lot simpler than what you tried doing. :smile:
 
  • #4
48
0
i manged to get it to [Sin2x+1+2cosx+cos2x] / SinxCosx


That should actually be [tex]\frac{sin^{2}x+1+2cosx+cos^{2}x}{sinx+sinxcosx}[/tex]

Simplify and then factor.
 
  • #5
63
0
Woohoo got it !! ..

Sin2x + cos2x will be 1

so ..

it will be [2 + 2cosx] / [sinx + sinxcosx]

Cosx cancels each other

==> 4/ 2sinx

==> 2/sinx

==> 2cscx

Thanks alot for the help guys :D
 
  • #6
63
0
I have anther question...

Find the general solutions for the following equations.

2 cos2 x -1 =0

Soultion.

Cos2x = 1/2
Cosx= - or + Sqrt 2 / 2

The general solution is pi /4 +2nPi, 3pi/4 +2nPi , 5Pi/4 + 2nPi and 7Pi/4 +2nPi , where n is an integer




But why can't i just take the Double angle of 2cos2x ?

which will be, cos2x=0

The general solution is Pi/2 +2nPi and 3Pi/2 +2nPi where n is an integer..



Btw, that was an example from my text book...the first solution is what the text book stated, and the second solution is what i thought right ..
 
  • #7
48
0
Woohoo got it !! ..

Sin2x + cos2x will be 1

so ..

it will be [2 + 2cosx] / [sinx + sinxcosx]

Cosx cancels each other

==> 4/ 2sinx

==> 2/sinx

==> 2cscx

Thanks alot for the help guys :D

Mspike, don't make that mistake. The cosx don't divide out. Since you have already pretty much attained your answer, this is what should have happened:

[tex]\frac{2+2cosx}{sinx+sinxcosx}[/tex]

[tex]=\frac{2(1+cosx)}{sinx(1+cosx)}[/tex] The (1+cosx) in the numerator and denominator divide out.

[tex]=\frac{2}{sinx}[/tex]

[tex]=2cscx[/tex]
 
  • #8
63
0
Mspike, don't make that mistake. The cosx don't divide out. Since you have already pretty much attained your answer, this is what should have happened:

[tex]\frac{2+2cosx}{sinx+sinxcosx}[/tex]

[tex]=\frac{2(1+cosx)}{sinx(1+cosx)}[/tex] The (1+cosx) in the numerator and denominator divide out.

[tex]=\frac{2}{sinx}[/tex]

[tex]=2cscx[/tex]

Ahh!.
thanks for the heads up, and i thought i knew it :S
 
  • #9
48
0
I have anther question...

Find the general solutions for the following equations.

2 cos2 x -1 =0

Soultion.

Cos2x = 1/2
Cosx= - or + Sqrt 2 / 2

The general solution is pi /4 +2nPi, 3pi/4 +2nPi , 5Pi/4 + 2nPi and 7Pi/4 +2nPi , where n is an integer




But why can't i just take the Double angle of 2cos2x ?

which will be, cos2x=0

The general solution is Pi/2 +2nPi and 3Pi/2 +2nPi where n is an integer..



Btw, that was an example from my text book...the first solution is what the text book stated, and the second solution is what i thought right ..

In your solution, you did not even mention pi/4. The double angle is as follows:
2cos[tex]^{2}[/tex]x-1=0
cos2x=0
2x=pi/2
x=pi/4

That is only one solution, and you are still missing it. Check where you went wrong.
 
  • #10
63
0
Am sorry for asking so many questions ,

how one should factor , tan2-sec-1=0

i think it will be something like, (x-1)(x2+1)

where x and x2 are trig
 
  • #11
rock.freak667
Homework Helper
6,223
31
Am sorry for asking so many questions ,

how one should factor , tan2-sec-1=0

i think it will be something like, (x-1)(x2+1)

where x and x2 are trig

try using an identity involving tan2x and sec2x to replace the tan2x term.
 
  • #12
63
0
Solve Tan2x-secx-1=0

Soultion

(Sec2x-1)-secx-1=0

sec2-secx-2=0


(secx-2)(secx+1)=0


--------------
Secx -2 =0
secx=2
x= Pi/3 , 5Pi/3

-------------
Secx+1=0
Secx=-1
x= Pi


X= Pi/3 , 5Pi/3, Pi


Is that right? am i missing something ?

Thank you :D
 
  • #13
rock.freak667
Homework Helper
6,223
31
Solve Tan2x-secx-1=0

Soultion

(Sec2x-1)-secx-1=0

sec2-secx-2=0


(secx-2)(secx+1)=0


--------------
Secx -2 =0
secx=2
x= Pi/3 , 5Pi/3

-------------
Secx+1=0
Secx=-1
x= Pi


X= Pi/3 , 5Pi/3, Pi


Is that right? am i missing something ?

Thank you :D

looks fine to me
 
  • #14
48
0
Solve Tan2x-secx-1=0

Soultion

(Sec2x-1)-secx-1=0

sec2-secx-2=0


(secx-2)(secx+1)=0


--------------
Secx -2 =0
secx=2
x= Pi/3 , 5Pi/3

-------------
Secx+1=0
Secx=-1
x= Pi


X= Pi/3 , 5Pi/3, Pi


Is that right? am i missing something ?

Thank you :D

Yup, if the domain was [0,2pi], then it is correct.
 

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