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PreCal Trig Identities Project. Working with 1 side.
- Thread starter dkenon
- Start date
fasteddie65
Full Member
- Joined
- Nov 1, 2008
- Messages
- 360
#3 sin ø (csc ø - sin ø) = cos^2 ø
sin ø (csc ø - sin ø) = 1 - sin^2 ø
The rest is easy.
sin ø (csc ø - sin ø) = 1 - sin^2 ø
The rest is easy.
Hello, dkenon!
\(\displaystyle \text{Factor: }\:\cos^2\!\theta\underbrace{\left(1 + \tan^2\!\theta\right)}_{\text{This is }\sec^2\!\theta} \;=\;\cos^2\!\theta \cdot\sec^2\!\theta \;=\;(\underbrace{\cos\theta\sec\theta}_{\text{This is 1}})^2 \;=\;1^2 \;=\;1\)
\(\displaystyle \text{As Loren suggested, factor: }\:\sec^2\!\theta\underbrace{(1 + \tan^2\!\theta)}_{\text{This is }\sec^2\!\theta} \:=\:\sec^2\!\theta\cdot\sec^2\!\theta \;=\;\sec^4\!\theta\)
\(\displaystyle \text{The left side is: }\:\underbrace{\sin\theta\csc\theta}_{\text{This is 1}} - \sin^2\!\theta \;=\;\underbrace{1 - \sin^2\!\theta}_{\text{This is }\cos^2\!\theta} \;=\;\cos^2\!\theta\)
\(\displaystyle 1)\;\;\cos^2\!\theta +\tan^2\!\theta\cos^2\!\theta \:=\:1\)
\(\displaystyle \text{Factor: }\:\cos^2\!\theta\underbrace{\left(1 + \tan^2\!\theta\right)}_{\text{This is }\sec^2\!\theta} \;=\;\cos^2\!\theta \cdot\sec^2\!\theta \;=\;(\underbrace{\cos\theta\sec\theta}_{\text{This is 1}})^2 \;=\;1^2 \;=\;1\)
\(\displaystyle 2)\;\;\sec^2\!\theta + \tan^2\!\theta\sec^2\!\theta \:=\sec^4\!\theta\)
\(\displaystyle \text{As Loren suggested, factor: }\:\sec^2\!\theta\underbrace{(1 + \tan^2\!\theta)}_{\text{This is }\sec^2\!\theta} \:=\:\sec^2\!\theta\cdot\sec^2\!\theta \;=\;\sec^4\!\theta\)
\(\displaystyle 3)\;\;\sin\theta(\csc\theta-\sin\theta) \:=\:\cos^2\!\theta\)
\(\displaystyle \text{The left side is: }\:\underbrace{\sin\theta\csc\theta}_{\text{This is 1}} - \sin^2\!\theta \;=\;\underbrace{1 - \sin^2\!\theta}_{\text{This is }\cos^2\!\theta} \;=\;\cos^2\!\theta\)