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
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
Use the fact that x^3 + y^3 = (x + y)(x^2 – xy + y^2)b) cos^3Θ + sin^3Θ = (cosΘ + sinΘ)(1 - sinΘcosΘ)
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!