Trigonometric Identities algebra

[Mspike6]

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

 

[berkeman]

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

There's an identity involving sin²x and cos²x that might help get you a bit farther...

 

[Bohrok]

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.

 

[Anakin_k]

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

That should actually be $$\frac{sin^{2}x+1+2cosx+cos^{2}x}{sinx+sinxcosx}$$

Simplify and then factor.

 

[Mspike6]

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 a lot for the help guys :D

 

[Mspike6]

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 textbook...the first solution is what the textbook stated, and the second solution is what i thought right ..

 

[Anakin_k]

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 a lot 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:

$$\frac{2+2cosx}{sinx+sinxcosx}$$

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

$$=\frac{2}{sinx}$$

$$=2cscx$$

 

[Mspike6]

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:

$$\frac{2+2cosx}{sinx+sinxcosx}$$

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

$$=\frac{2}{sinx}$$

$$=2cscx$$

Ahh!.
thanks for the heads up, and i thought i knew it :S

 

[Anakin_k]

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 textbook...the first solution is what the textbook 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$$^{2}$$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.

 

[Mspike6]

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

 

[rock.freak667]

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.

 

[Mspike6]

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

 

[rock.freak667]

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

 

[Anakin_k]

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.