Stuck On Identities.

[MJFan]

So, I have this problem:

. . . . .\(\displaystyle \sin(x)\, \left(\tan(x)\right)\, +\, \dfrac{\sin(x)}{\tan(x)}\, =\, \dfrac{1}{\cos(x)}\)

and I can get to here no problem:

. . . . .\(\displaystyle \sin(x)\, \left(\tan(x)\right)\, +\, \dfrac{\sin(x)}{\tan(x)}\, =\, \dfrac{1}{\cos(x)}\)

. . . . .\(\displaystyle \mbox{Use LS}\)

. . . . .\(\displaystyle =\, \sin(x)\, \left(\dfrac{\sin(x)}{\cos(x)}\right)\, +\, \dfrac{\sin(x)}{\tan(x)}\)

. . . . .\(\displaystyle =\,\cos(x)\,+\,\dfrac{\sin(x)}{\left(\dfrac{\sin(x)}{\cos(x)}\right)}\)

. . . . .\(\displaystyle =\, \cos(x)\, +\, \left(\, \dfrac{\sin(x)}{1}\, \cdot\, \dfrac{\cos(x)}{\sin(x)}\, \right)\)

. . . . .\(\displaystyle =\, 2\cos(x)\)

But I don't know what to do from here. Any help is appreciated!

 

[stapel]

So, I have this problem:

. . . . .\(\displaystyle \sin(x)\, \left(\tan(x)\right)\, +\, \dfrac{\sin(x)}{\tan(x)}\, =\, \dfrac{1}{\cos(x)}\)

What are the instructions for this exercise? Are you supposed to be solving an equation, or proving an identity, or something else?

and I can get to here no problem:

. . . . .\(\displaystyle \sin(x)\, \left(\tan(x)\right)\, +\, \dfrac{\sin(x)}{\tan(x)}\, =\, \dfrac{1}{\cos(x)}\)

. . . . .\(\displaystyle \mbox{Use LS}\)

What does "Use LS" mean? Is "LS" a theorem or algorithm you were given? Or does this mean "operate only on the left-hand side (LHS) of the equation"?

. . . . .\(\displaystyle =\, \sin(x)\, \left(\dfrac{\sin(x)}{\cos(x)}\right)\, +\, \dfrac{\sin(x)}{\tan(x)}\)

. . . . .\(\displaystyle =\,\cos(x)\,+\,\dfrac{\sin(x)}{\left(\dfrac{\sin(x)}{\cos(x)}\right)}\)

How did you go from:

. . . . .\(\displaystyle \sin(x)\, \left(\dfrac{\sin(x)}{\cos(x)}\right)\, =\, \left(\, \dfrac{\sin(x)}{1}\,\right)\, \left(\, \dfrac{\sin(x)}{\cos(x)}\,\right)\, =\, \dfrac{\sin^2(x)}{\cos(x)}\)

...to:

. . . . .\(\displaystyle \cos(x)\)

...as the first term on the left-hand side?

 

[MJFan]

What are the instructions for this exercise? Are you supposed to be solving an equation, or proving an identity, or something else?


What does "Use LS" mean? Is "LS" a theorem or algorithm you were given? Or does this mean "operate only on the left-hand side (LHS) of the equation"?


How did you go from:

. . . . .\(\displaystyle \sin(x)\, \left(\dfrac{\sin(x)}{\cos(x)}\right)\, =\, \left(\, \dfrac{\sin(x)}{1}\,\right)\, \left(\, \dfrac{\sin(x)}{\cos(x)}\,\right)\, =\, \dfrac{\sin^2(x)}{\cos(x)}\)

...to:

. . . . .\(\displaystyle \cos(x)\)

...as the first term on the left-hand side?

Sorry, I should have explained better. My teacher wants LS to show that you are going to use the left side of the equation, and yes I am trying to prove that both sides are equal, I cancelled out the (sinx)'s. But now that I am looking at it again, I don't think you could cancel those. :S

 

[stapel]

What does "Use LS" mean? ...does this mean "operate only on the left-hand side (LHS) of the equation"?

Sorry, I should have explained better. My teacher wants LS to show that you are going to use the left side of the equation...

Thank you for the clarification.

How did you go from:

. . . . .\(\displaystyle \sin(x)\, \left(\dfrac{\sin(x)}{\cos(x)}\right)\, =\, \left(\, \dfrac{\sin(x)}{1}\,\right)\, \left(\, \dfrac{\sin(x)}{\cos(x)}\,\right)\, =\, \dfrac{\sin^2(x)}{\cos(x)}\)

...to:

. . . . .\(\displaystyle \cos(x)\)

...as the first term on the left-hand side?

...I cancelled out the (sinx)'s. But now that I am looking at it again, I don't think you could cancel those.

Since both sines are in the numerators of their respective fractions, no, I don't see any way that they could be "cancelled". So what have you done other than that erroneous step? After you simplified correctly and then converted the two fractions on the left-hand side (LHS) to a common denominator, what did you get?

Please show all of your steps. Thank you!

 

[MJFan]

Thank you for the clarification.


Since both sines are in the numerators of their respective fractions, no, I don't see any way that they could be "cancelled". So what have you done other than that erroneous step? After you simplified correctly and then converted the two fractions on the left-hand side (LHS) to a common denominator, what did you get?

Please show all of your steps. Thank you!

If you have anything better to use than "use LS", it would be very helpful to know!
Btw, yes I know I spelled it wrong, it's just that I don't know how to go back and change it. :S

Which of course shows that both sides are equal to each other.