Online Course Discussion Forum
MCIV Trig Packet Questions
Hello,
For problem 6.14: I tried using the conclusion from part A. I then multiplied the numerator and denominator of each fraction by a sin(x), sin(A), sin(B), or sin(c), depending on the fraction. This way, I was able to use the doubel angle formula on the denominator and get sin^2x, sin^2A, sin^2B, and sin^C as the numerators, which is very similar to the format of the "want-to-show" equation. I would like some advice on what to do next/how to finish the problem.
For problem 6.15, I used sum to product on the cos(alpha)+cos(beta). I used the double angle formula to make cos(alpha+beta) to cos^2((alpha+beta)/2)-1. Then, I tried completing the square by seperating the 1 and the -2cos^2((alpha+beta)/2) into -cos^2((alpha+beta)/2)-cos^2((alpha+beta)/2). I wasn't able to get a nice sum of squares that equals 0; I got a sum of squares equal to 1/2. What do I do from here?
For problem 6.17 part a: I started with the 4sinAsinBsinC first. 4sinAsinBsinC=4sinAsinBsin(A+B)=-2(cos(A+B)-cos(A-B))sin(A+B)=2cos(A-B)cos(A+B)-2cos(A+B)sin(A+B)=2cos(A-B)cos(A+B)-sin(2A+2B). LHS=sin2A+sin2B+sin(2A+2B). If LHS=RHS, sin2A+sin2B+sin(2A+2B)=2cos(A-B)cos(A+B)-sin(2A+2B). sin2A+sin2B+sin(2A+2B)=cos(90-2A)=cos(90-2B)=cos(90-2A-2B), then move them to the RHS and use sum to product to get -2sin(45)sin(2A)-2sin(45)sin(2B)-2cos(45)cos(2A+2B)=0. Cancelling out terms, sin(2A)+sin(2B)+cos(2A+2B)=0. How do I prove this is true? I would also like to have some advice for parts b and c. I don't really know how to do those either.
For problem 6.19, I tried drawing the picture, and writing out tan(A/2)=r/x, tan(B/2)=r/z, and tan(C/2)=r/y and working with that, but I didn't seem to any progress from there. (x, y, and z are the lengths on sides a, b, c that are from the tangent point of the incircle with the side to a vertex on that side). How do I procceed?
Thank you!
Tina Jin
Let me respond one at a time.
For 6.14(b), Use the result of part (a): $\cot X = \cot A + \cot B + \cot C$, and then use the identity $\cot^2 x + 1 = \csc^2 x$. Square both sides and add $1$. Then you will use the result for Example 7: $\cot A \cot B + \cot B\cot C + \cot C\cot A = 1$ to get the result.
For 6.15, your approach is correct, but I am not sure you had typos or you used the identities incorrectly. In any case, try to apply the correct identities very carefully, and follow the method of Problem 6.9 for completing the square.
6.17 (a): This is the same question as in Example 6. There is an error in your algebra.
Part (b): Use the Product-to-Sum formulas for each term, and remember $A+B+C=\pi$.
Part (c)&(d): These two are equivalent (can you prove it?). The original problem is just the part (d). I split it into four parts so you can work out (a)(b)(c) and then use them in (d).
To prove (c), use the triple angle formula to get the following:
$$\sin^3 A = \dfrac{1}{4}(3\sin A - \sin 3A).$$
In addition to Part (b), you also need another identity:
$$\sin A\cos(B-C) + \sin B\cos(C-A) + \sin C\cos(A-B) = 4\sin A\sin B\sin C.$$
The strategy: you may handle each term separately, and then use the symmetry to get the other terms.
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