Simplifying trig expressions (help please)

[dannyp32]

i need help with these problems
i really dont understand how to do them
1.
[sin(x)-sin(x)cos(x)]/[1-cos(x)]

2.
[2sin (x)]/[1-cos^2(x)]


i really appreciate your help

 

[Deleted member 4993]

i need help with these problems
i really dont understand how to do them
1.
[sin(x)-sin(x)cos(x)]/[1-cos(x)] <<< Hint: factorize the numerator - what is the common factor?

2.
[2sin (x)]/[1-cos^2(x)] <<< Hint: Use .... Sin[sup:1os642uj]2[/sup:1os642uj]x + Cos[sup:1os642uj]2[/sup:1os642uj]x = 1


i really appreciate your help

 

[dannyp32]

thanks for your quick reply

sorry but i dont know how for number 1

*edited: do you mean: [sin x (1-cos(x)]/[1-cos(x)]
and then 1-cosx cancels out so the answer is sin x


#2 yeah i did that and i got [2sin(x)]/[sin^2(x)]
but im stuck there

 

[Loren]

i need help with these problems
i really dont understand how to do them
1.
[sin(x)-sin(x)cos(x)]/[1-cos(x)] <<< Same as \(\displaystyle \frac{a-ab}{1-b}\). Can you simplify that?

2.
[2sin (x)]/[1-cos^2(x)] <<< Hint: 1-cos[sup:2jp3tk91]2[/sup:2jp3tk91]x = sin[sup:2jp3tk91]2[/sup:2jp3tk91]x. Then same as \(\displaystyle \frac{2a}{a^2}\).


i really appreciate your help

 

[Aladdin]

thanks for your quick reply

sorry but i dont know how for number 1

*edited: do you mean: [sin x (1-cos(x)]/[1-cos(x)]
and then 1-cosx cancels out so the answer is sin x-->yes , The result is sin(x) over 1


#2 yeah i did that and i got [2sin(x)]/[sin^2(x)]--> \(\displaystyle \frac{2}{sin(x)}\)
but im stuck there

 

[Loren]

>>i need help with these problems
i really dont understand how to do them

You then posted two expressions. You did not include the instructions as to what you are supposed to do with them. It would be helpful if you included the instructions that you were given, word for word.