How do I solve these two trigonometric identifies?

Question

Hello, I'm kind of stuck with these two.

[sin^2(x)-2sin(x)+1] / [sin(x)-1] = sin(x)-1

The other:

[ sin(x) / 1-cos(x) ] - [ sin(x) / 1+cos(x) ] = 2cot(x)

If you could go step by step so I can understand where you're coming from, that would be much apperciated.

Many thanks.

Answer 1

For the first one.

1) [sin^2(x)-2sin(x)+1] / [sin(x)-1] = sin(x)-1

Factorize the numerator on the left hand side of the equation:

2) [(sin(x)-1)^2] / [sin(x)-1] = sin(x)-1

If this step gave you trouble, think of y as being equal to sin(x), so y = sin(x). Then the numerator becomes y^2-2y+1. Factoring this gives us: (y-1)^2. Substituting in back sin(x), we get that the numerator is equal to (sin(x)-1)^2.

Now, since the numerator and the denominator on the left hand side both have a common multiple (sin(x)-1), we can divide, leaving us with:

3) sin(x) -1 = sin(x) -1

----------------------------------------------------

For the second one

1) [ sin(x) / 1-cos(x) ] - [ sin(x) / 1+cos(x) ] = 2cot(x)

The first step is to find a common denominator for the left hand side. If we multiply the denominators of the 2 terms on the left hand side, we get:

2) [sin(x)*(1+cos(x)) - sin(x)*(1-cos(x))] / [1 - cos^2(x)] = 2cotx

Note: "*" is the same as "multiply". And 1-cos^2(x) = (1+cosx)(1-cosx) [formula for difference of two squares].

Now, we can eliminate terms on the numerator for the left hand side. For one, the sin(x)'s cancel out, leaving us with:

3) [2sin(x)cos(x)] / 1-cos^2(x)] = 2 cotx

Divide both sides by 2. This will leave us with:


4) [sin(x)cos(x)] / 1-cos^2(x)] = cotx

Now, in order to go further, we need to first find another identity. We know that sin^2(x) + cos^2(x) = 1. This can be manipulated so we get: sin^2(x) = 1 - cos^2(x).

Thus: by substituting 1-cos^2(x) with sin^2(x) back into the equation, we get:

5) sin(x)cos(x) / sin^2(x) = cot(x)

Simplifying gives us:

6) cos(x) / sin(x) = cot(x)

This of course is an identity, as cos(x)/sin(x)=cot(x)!!

Thus, we get:

7) cot(x) = cot(x).

If you have anymore questions on this, feel free to e-mail me.

Answer 2

Are you trying to prove the identities?

[sin^2(x)-2sin(x)+1] / [sin(x)-1]
= (sin x - 1)^2 / (sin x - 1)
= sin x - 1

[ sin(x) / 1-cos(x) ] - [ sin(x) / 1+cos(x) ]
Get a common denom (1 - cos x)(1 + cos x)
= [(sin x)(1+cos x) - (sin x)(1-cos x)] / (1-cos x)(1+cos x)
= (sin x + sin x cos x - sin x + sin x cos x) / (1 - cos^2 x)
= 2 sin x cos x / (1 - cos^2 x)
= 2 sin x cos x / sin^2 x
= 2 cos x / sin x
= 2 cot x

Answer 3

[sin^2(x)-2sin(x)+1] / [sin(x)-1] = sin(x)-1

If we say y=sin(x), the equation is rewritten as

(y^2 - 2y + 1)/(y - 1) = (y - 1)

We can factor (y^2 - 2y + 1) to be (y - 1)^2

So, we have

((y - 1)^2)/(y - 1) = (y - 1)

Or, (y - 1) = (y - 1)

-----------------------------------------------

[ sin(x) / 1-cos(x) ] - [ sin(x) / 1+cos(x) ] = 2cot(x)

If we combine the fractions in the left side we get:

(sin(x) * (1+cos(x)) - sin(x) * (1-cos(x))) / (1 - cos^2(x)) = 2cot(x)
(sin(x) + cos(x)sin(x) - sin(x) + cos(x)sin(x)) / (1 - cos^2(x)) = 2cot(x)
2cos(x)sin(x) / (1 - cos^2(x)) = 2cot(x)
cos(x)sin(x) / (1 - cos^2(x)) = cot(x)

Remembering that cotangent is the cosine devided by the sine ...

cos(x)sin(x) / (1 - cos^2(x)) = cos(x)/sin(x)
sin(x) / (1 - cos^2(x)) = 1/sin(x)

Dividing both sides by sin(x):

1 / (1 - cos^2(x)) = 1 / sin^2(x)

Inverting both sides we get:

1 - cos^2(x) = sin^2(x)

or ...

sin^x(x) + cos^2(x) = 1 (which is an accepted fact)

QED

Answer 4

Question 1
(sin x - 1) (sin x - 1) / (sin x - 1)
(sin x) - 1

Question 2
LHS = N/D
D = (1 - cos x)(1 + cos x):-
N = [sin x (1 + cos x) - sin x(1 - cos x)
N = sin x + sin x cos x - sin x + sin x cos x
N = 2 sin x cos x
N / D = 2 sin x cos x / (1 - cos x) (1 + cos x)
N / D = 2 sin x cos x / (1 - cos ² x)
N / D = 2 sin x cos x / (sin ² x)
N / D = 2 cot x

Answer 5

1)The numerator is of the form a^2+2ab+b^2.
Which is equal to (a+b)^2
Therefore numerator of(1) =[sin^2(x)-2sin(x)+1] =[sin(x)-1]^2
Thus G.E= [(sin(x)-1)^2]/[sin(x)-1]
so now denominator cancels with the one term in the numerator
Thus left with [sin(x)-1].

2)L.H.S=[ sin(x) / 1-cos(x) ] - [ sin(x) / 1+cos(x) ]
First take the L.C.M
so the denominator= [1-cos(x)][1-cos(x)]
which is in the form (a+b)(a-b)=a^2-b^2
So denominator=[1^2-cos^2(x)]=[1-cos^2(x)]
Now we know that sin^2(x)+cos^2(x)=1
So1-cos^2(x)=sin^2(x)
And after taking the L.C.M the numerator is
[sin(x){1+cos(x)}]-[ sin(x) {1-cos(x) }]
=[sin(x)+sin(x)cos(x)]-[sin(x)-sin(x)cos(x)]
=[sin(x)+sin(x)cos(x)-sin(x)+sin(x)cos(x)]
=2sin(x)cos(x)
Thus G.E=2sin(x)cos(x)/sin^2(x)
=2cos(x)/sin(x)
=2cot(x).

Answer 6

[sin^2(x)-2sin(x)+1] / [sin(x)-1] =
=[sin(x)-1]^2/ [sin(x)-1] = sin(x)-1

[ sin(x) / 1-cos(x) ] - [ sin(x) / 1+cos(x) ] =
=sin(x)*{[1/(1-cosx)]-
-[1/1+cos(x)]}=
=sinx*(2cosx/sin^2x)=2ctgx