Expert: Scott A Wilson - 7/28/2009

Question
Having some difficulty with these problems and don't have a teacher to ask to explain them and show me step by step how to do them. Please show your work and explain how you got the answer .Your help is much appreciated.

1. Express cot² x-1/ csc² in terms of sin x

2. Verify the following identities.

a. (sin B + cos B)(sin B - cos B)=2 sin² B-1

b. (1-cos²(y) + sin²(y))² +4 sin²(y) cos²(y)= 4 sin²(y)

c. tan²Θ sec²Θ - sec²Θ + 1 = tan^4Θ

d. sinΘ / cscΘ + cosΘ / secΘ = 1

e sin(x) tan²(x) cot³(x) = cos(x)

f. sec² X / sec² X-1 = csc² X

                          Thanks in advance,
                                           Anthony

Answer
Now normally, I don't answer a question that requires this many answers.
However, I only have two other people asking me questions,
so I'll answere this one as well.

1. Express cot² x-1/ csc²x in terms of sin x
It is known that cot = cos/sin, and cos=√(1-sin²).
It is also known that csc=1/sin.
Since there is no speace by the x-1, the way the problem looks is is cot²(x-1)/csc²x.

However, there also needs to be a space after the 1.
The problem could also be put as (cot²x - 1) / csc²x if the 1 is not included in the cot().

I will choose (cot²x - 1) / csc²x.

Now cot²x + 1 = csc²x, so cot²x = csc²x - 1,
so that problem could be seen as (csc²x - 2) / csc²x.

Knowing that cscx = 1 / sinx, that converts to (1/sin²x - 2)/(1/sin²x).

Multiplying the top and bottom by sin²x gives us 1 - 2sin²x.


2. Verify the following identities.
a. (sin B + cos B)(sin B - cos B) =2 sin² B-1
Working on the right side and changing it into the left, we get the following:
Note that sin²x + cos²x = 1, so sin²B – 1 = -cos²B.
This means that 2 sin²B – 1 = sin²B - cos²B.
This is the difference between two squares and factors into (sin B + cos B)(sin B - cos B).

b. (1 - cos²(y) + sin²(y))² + 4sin²(y)cos²(y)= 4 sin²(y)
Using the left side, I’ll get the right side.
First, since sin²y + cos²y = 1, 1 – cos²y = sin²y.
This means that (1 - cos²(y) + sin²(y))² + 4sin²(y)cos²(y) is really (2sin²(y))² + 4sin²(y)cos²(y).
Squaring the first term gives us 4sin²(y)sin²(y) + 4sin²(y)cos²(y).
The sin²(y) was written twice to make known it is both terms and along with the 4, can be factored out.
This gives 4sin²(y)(sin²(y)+cos²(y)) = 4sin²(y)(1) =4 sin²(y).

c. tan²Θ sec²Θ - sec²Θ + 1 = tan^4Θ
On this one, it trig problems, there might be a better way to do it, but I find it easiest to convert to sin() and cos() if the answer doesn’t appear to me.
Taking the left side and turning it to the right side gives (sin²Θ/cos²Θ)(1/cos²Θ) – 1/cos²Θ + 1.
Putting the trig values all over cos^4(Θ) gives (sin²Θ – cos²Θ + cos^4(Θ)/cos^4(Θ).
Replacing cos²Θ out of cos^4(Θ) with 1-sin²Θ gives (sin²Θ - cos²Θ + (1-sin²Θ)cos²Θ)/cos^4Θ.
Multiplying the top out gives (sin²Θ - cos²Θ + cos²Θ - sin²Θcos²Θ)/cos^4Θ.
Canceling terms in the middle gives (sin²Θ - sin²Θcos²Θ)/cos^4Θ.
Factoring out sin²Θ in the numerator gives sin²Θ(1 - cos²Θ)/cos^4Θ.
Since we know that 1 – cos²Θ = sin²Θ, this gives sin^4(Θ)/cos^4(Θ).
Since sinΘ/cosΘ = tanΘ, this is tan^4(Θ).

d. sinΘ / cscΘ + cosΘ / secΘ = 1
Since cscΘ =1/sinΘ and secΘ = 1/cosΘ, the left side becomes sin²Θ + cos²Θ, and we all know that is 1.

e sin(x) tan²(x) cot³(x) = cos(x)
Putting tanx as sinx/cosx and cotx as cosx/sinx, we get sinx(sinx/cosx)^2(cosx/sinx)^3 =
sinx(cosx/sinx) = cosx.

f. sec² X / sec² X-1 = csc² X
It is known from the identities that tan²X +  1 = sec²X, so sec²X – 1 = tan²X.
It is also known that tanX = sinX / cosX and secX = 1 / cosX, so we have (1/cos²X)/(sin²X/cos²X).
Invert the denominator and multiply and we get (1/cos²X)(cos²X / sin²X).
Since cos²X/cos²X is 1, this is 1/sin²X, which is csc²X since 1/sinX is cscX.