How Do You Solve Complex Trigonometric Problems?

AI Thread Summary
To solve complex trigonometric problems, understanding the unit circle and trigonometric identities is essential. For the first problem, given sin(theta) = 1/4 in quadrant II, the cosine can be calculated using the cosine of a sum identity. The second problem involves simplifying the sum of inverse sine and cosine functions, which may require a calculator for exact values. The third problem can be approached using sine and cosine difference identities for simplification. Mastery of these concepts can significantly aid in tackling trigonometric challenges effectively.
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1. Hi, I have a pre-calculus final tomorrow, and there are a few questions I don't understand.
I'd truly appreciate it if you could help :)

1. If sin(theta)=1/4, theta in quadrant II, find the exact value of cos(theta+pi/6)

2. sin(sin^-1(2/3) + cos^-1(1/3)) -- simplify

3. cos(5pi/18)sin(pi/9) - cos(pi/9)sin(5pi/18) -- simplify


2. n/a



3.

1. if sin theta = 1/4, that would mean the sides are 1, 4, root15. Use coscos-sinsin (composite argument) and plug into make [(root 15)/4 * root3/2 ] - [1/4 * 1/2] then...?

2. I'm truly confused about this one... do you use the calculator to find the individual inverse sine/cosine functions? then add the two, and use the sin function on them? If so, do you use the graphing calculator?

3. this is the sine composite function (subtracting), I know that, but I do not know how to continue on...




I am so bad at these, and I'm freaking out for my final... :(

Thank you!
 
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θ is in Quadrant II, so cos(θ) is negative.
 
For #3, just take a look at the sum/difference identities:

Once you have it in the form sin(x-y), it should be easy to simplify..
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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