Pre-calc project
Texte du Storyboard
- Jackson's School Day
- By Iain Patterson
- It seems like a lot but it is fairly simple. First, substitute one of trig identities. In this case you can change cos^2x to 1-sin^2x because of the Pythagorean Identity. This will then create the equation -3sin^2x-[(1-sin^2x)sin^2x]+1-cos^2x=0
- After doing this, you need to do what is in the parentheses first. This gets you -3sin^2x-sin^2x+sin^4x+1-sin^2x=0. You can now combine like terms to simplify. You get -5sin^2x+sin^4x+1=0
- With Identities means you have trig identities as a part of the equation. For example, the equation -3sin^2x-cos^2xsin^2x+cos^2x=0for (0,2pi]
- You can now factor out a sin^2x which leaves you with sin^2x(-5+sin^2x+1), you combine like terms in the parentheses and get sin^2x(sin^2x-4). Set both things equal to 0. Sin^2x=0, square root and get sinx=0.
- You are left with sin^2x-4=0. Add the 4 and square root. you get sinx=+-2. This answer is outside the range so it does not work. sinx=0 does however because this gets 3 answers, those being 0,pi, and 2pi. Due to the restriction we can only use the last two.
- Oooo, Fancy. What is with identities?
- So the answer is pi and 2pi
- Yeah! We learned about solving trig functions with and without identities.
- If you had the equation 4sin^2x-3=0 the first thing you would need to do is add the three over to the 0. After that you get 4sin^2x=3. You then divide by 4 which gets you sin^2x=3/4. To get rid of the square, you need to square root! After doing that you get sinx= +- (sqrt3)/2
- Anything else you learned that was intersting?
- This gives multiple solutions on the unit circle which include pi/3, 2pi/3, 4pi/3, and 5pi/3
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