Thread: proving trig identities

a) 1/cos^2Θ + 1/sin^2Θ = (tanΘ + 1/tanΘ)^2

b) cos^4Θ - sin^4Θ = cos^2Θ - sin^2Θ


thank youu.

Originally Posted by checkmarks

b) cos^4Θ - sin^4Θ = cos^2Θ - sin^2Θ
.

Difference of two square

(cos^2 x+sin^2 x)(cos^2 x - sin^2 x)

Pythagoren Identity

cos^2 x - sin^2 x

Hello, checkmarks!

a) .1/cos²θ + 1/sin²θ .= .(tanθ + 1/tanθ)²

This is a messy one . . . Start with the right side.

. . . . . . . . .sinθ . . cosθ . . . . . . sin²θ + cos²θ . . . . . . . . . 1
We have: .(------ + ------)² . = . (-----------------)² . = . --------------
. . . . . . . . .cosθ . .sinθ . . . . . . . . sinθ·cosθ . . . . . . . sin²θ·cos²θ

. . . . .sin²θ + cos²θ . . . . . .sin²θ . . . . . . .cos²θ . . . . . . . .1 . . . . . 1
. . = . ---------------- . = . -------------- + --------------- . = . ------- + -------
. . . . . .sin²θ·cos²θ . . . . sin²θ·cos²θ . . sin²θ·cos²θ . . . . .cos²θ . . sin²θ

thanks!!

i also need help with these 3:

a) tan^2Θ - sin^2Θ = tan^2Θ(sin^2Θ)

b) cos^3Θ + sin^3Θ = (cosΘ + sinΘ)(1 - sinΘcosΘ)

Originally Posted by checkmarks

a) tan^2Θ - sin^2Θ = tan^2Θ(sin^2Θ)

Consider LHS
tan^2Θ - sin^2Θ = sin^2 Θ/cos^2 Θ – sin^2 Θ
…………………= (sin^2 Θ – cos^2 Θsin^2 Θ)/cos^2 Θ
…………………= [sin^2 Θ(1 – cos^2 Θ)]/cos^2 Θ
…………………= (sin^2 Θ * sin^2 Θ)/cos^2 Θ
…………………= sin^2 Θ * sin^2 Θ/cos^2 Θ
…………………= sin^2 Θ * tan^2 Θ
…………………= RHS

Thus we see, LHS = RHS

b) cos^3Θ + sin^3Θ = (cosΘ + sinΘ)(1 - sinΘcosΘ)

Use the fact that x^3 + y^3 = (x + y)(x^2 – xy + y^2)

Consider LHS
cos^3Θ + sin^3Θ = (cos Θ + sin Θ)(cos^2 Θ - sin Θcos Θ + sin^2 Θ) …sin^2 Θ + cos^2 Θ = 1
………………….= (cos Θ + sin Θ)(1 - sin Θcos Θ)
………………….= RHS

Thus we see, LHS = RHS

OR

If you didn’t remember the rule x^3 + y^3 = (x + y)(x^2 – xy + y^2), it would be a little harder

b) cos^3Θ + sin^3Θ = (cosΘ + sinΘ)(1 - sinΘcosΘ)
Consider RHS
(cosΘ + sinΘ)(1 - sinΘcosΘ) = cosΘ – sinΘcos^2Θ + sinΘ – sin^2ΘcosΘ
……………………………...= cosΘ(1 – sinΘcosΘ) + sinΘ(1 – sinΘcosΘ)
……………………………...= cosΘ(sin^2Θ + cos^2Θ – sinΘcosΘ) + sinΘ(sin^2Θ + cos^2Θ – sinΘcosΘ)
……………………………...= cosΘsin^2Θ + cos^3Θ – sinΘcos^2Θ + sin^3Θ + sinΘcos^2Θ – sin^2ΘcosΘ
……………………………...= cos^3Θ + sin^3Θ
...........................= LHS
Thus we see, LHS = RHS

Tips for approaching these kinds of problems:

1) Always start on the most complicated side
2) Change everything to sine and cosine. These are the trig functions all students are most familiar with, it will be easy to see patterns with them
3) Note that you may have to work on both sides. Working on one side and trying to get the other side does not always work. If you find you can't get the other side, stop and work on the other side to get where you left off on the first side.

Hello again, checkmarks!

a) .cscθ(cscθ + cotθ) .= .1/(1 - cosθ)

. . . . . . . . . 1. . . . .1 . . . cosθ . . . . .1 + cosθ . . . . 1 + cosθ
We have: . ----- · (------ + ------) . = . ---------- . = . ------------
. . . . . . . . sinθ . . sinθ . . sinθ . . . . . . sin²θ . . . . . 1 - cos²θ


. . . . . . . . . . . . . . . . . . .1 + cosθ . . . . . . . . . .1
Factor and reduce: . ------------------------ . = . ----------
. . . . . . . . . . . . . . .(1 - cosθ)(1 + cosθ) . . . .1 - cosθ

b) .tan²θ - sin²θ .= .tan²θ·sin²θ

The right side is: .tan²θ(1 - cos²θ) . = . tan²θ - tan²θ·cos²θ

. . . . . . . . . . . .sin²θ
. . . = . tan²θ - --------·cos²θ . = . tan²θ - sin²θ
. . . . . . . . . . . .cos²θ

c) .cos³θ + sin³θ .= .(cosθ + sinθ)·(1 - sinθ·cosθ)

The left side is the sum of two cubes . . . factor it!

. . . cos³θ + sin³θ . = . (cosθ + sinθ)·(cos²θ - sinθ·cosθ + sin²θ)


But sin²θ + cos²θ .= .1

. . Hence, we have: .(cosθ + sinθ)·(1 - sinθ·cosθ)


Hmmm, too slow . . . again!

Originally Posted by Soroban

Hmmm, too slow . . . again!

Sorry Soroban. As usual, I did not realize you were working on the problem, or I would have left it to you--you can give a lot more help than I can.

thanks so much, both of you
ive been given around 40-50 of these problems for the weekend, these are just the ones im a little stuck on

Originally Posted by checkmarks

thanks so much, both of you
ive been given around 40-50 of these problems for the weekend, these are just the ones im a little stuck on

how can you be a little stuck, either you're stuck or you're not

Originally Posted by Jhevon

how can you be a little stuck, either you're stuck or you're not

haha hey, give me a little slack
i usually start off correctly, i just dont know how to continue..hence..."a little stuck"

Originally Posted by checkmarks

haha hey, give me a little slack
i usually start off correctly, i just dont know how to continue..hence..."a little stuck"

ok, i'll accept that