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CM_Tutor

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I was asked to post some more questions. The following are in no particular order, and focus on no particular topic, and vary in standard...

1. ABC is any triangle. Show that
(a) tan A + tan B + tan C = tan A * tan B * tan C
(b) sin A + sin B + sin C = 4cos(A / 2)cos(B / 2)cos(C / 2)

2. For what positive integers m and n, if any, is the definite integral ∫ (0 --> 2π) sin mx * cos nx dx non-zero. If any non-zero values exist, what are they?

3. The lengths of the sides of a triangle form an arithmetic progression and the largest angle of the triangle exceeds the smallest by 90°. Find the ratio of the lengths of the sides.

4(a). By first simplifying sin(A + B) + sin(A - B), show that 2sin x(cos 2x + cos 4x + cos 6x) = sin 7x - sin x
(b) Hence, show that cos(π / 7) + cos(3π / 7) + cos(5π / 7) = 1 / 2

5. If b and c are positive integers, and a = b + c
(a) Use the binomial; theorem to show that a<sup>n</sup> - b<sup>n-1</sup>(b + cn) is divisible by c<sup>2</sup>
(b) Hence, show that 5<sup>42</sup> - 2<sup>48</sup> is a multiple of 9

6. A sequence is defined by u<sub>1</sub> = 12, u<sub>2</sub> = 30, and u<sub>n</sub> = 5u<sub>n-1</sub> -6u<sub>n-2</sub> for n &ge; 3. Use induction to prove that
u<sub>n</sub> = 2 * 3<sup>n</sup> + 3 * 2<sup>n</sup> for integers n &ge; 1.

7. Two circles, C<sub>1</sub> and C<sub>2</sub>, intersect at points K and L. KM is a tangent to circle C<sub>2</sub> and KN is a tangent to circle C<sub>1</sub>, where M lies on C<sub>1</sub> and N lies on C<sub>2</sub>. NL produced meets KM at P and circle C<sub>2</sub> at Q.
(a) Show that KN || MQ
(b) Show that &Delta;KMQ is isosceles

8. A tower AB and a building BC are located on level ground. A and C are the feet of the tower and the building, respecively, and CD is due south of AB. AB, of height H, is taller than CD, of height h, and the angle of depression of the top of CD from the top of AB is &beta;. E is a point on the ground, due east of tower AB. E is a distance d from A and 2d from C, and the angles of elevation of B and D, the tops of the tower and the building, from E are 2&alpha; and &alpha;, respectively.
(a) Draw a diagram to represent this information.
(b) Show that &radic;3 * tan &beta; = tan 2&alpha; - 2tan &alpha;
(c) If tan<sup>2</sup>&alpha; = tan &beta;, show that 2H = 3h.

9. Solve for x: sin<sup>-1</sup>x - cos<sup>-1</sup>x = sin<sup>-1</sup>(1 - x)

10. Suppose that z = cos 2&theta; + isin 2&theta;
(a) Show that (1 + z)<sup>n</sup> = 2<sup>n</sup>cos<sup>n</sup>&theta;(cos n&theta; + isin n&theta;)
(b) Use the binomial theorem to prove that &Sigma; (k=0 --> n) <sup>n</sup>C<sub>k</sub>cos 2k&theta; = 2<sup>n</sup>cos<sup>n</sup>&theta;cos n&theta;
(c) Hence, or otherwise, evaluate &int; (0 --> &pi; / 2) cos<sup>n</sup>&theta;cos n&theta; d&theta;

11(a). Use De Moivre's Theorem to show that cos 6&theta; = 32cos<sup>6</sup>&theta; - 48cos<sup>4</sup>&theta; + 18cos<sup>2</sup>&theta; - 1
(b) Hence, find an expression for the six roots of 32x<sup>6</sup> - 48x<sup>4</sup> + 18x<sup>2</sup> - 1 = 0
(c) Find the value of cos(&pi; / 12)cos(5&pi; / 12) and cos<sup>2</sup>(&pi; / 12) + cos<sup>2</sup>(5&pi; / 12)
(d) Hence, show that the roots of 32x<sup>6</sup> - 48x<sup>4</sup> + 18x<sup>2</sup> - 1 = 0 are +/-1 / &radic;2 and +/-(&radic;3 +/- 1) / 2&radic;2

12(a). If x and y are two positive real numbers, prove that x + y &ge; 2&radic;(xy)
(b) Suppose that &alpha; and &beta; are the two roots of x<sup>2</sup> + px = 1 / (2p<sup>2</sup>), where p is a non-zero real.
(i) Show that &alpha;<sup>4</sup> + &beta;<sup>4</sup> = (&alpha;<sup>2</sup> + &beta;<sup>2</sup>) / (2p<sup>2</sup>) - p(&alpha;<sup>3</sup> + &beta;<sup>3</sup>)
(ii) Hence, show that &alpha;<sup>4</sup> + &beta;<sup>4</sup> = 2 + p<sup>4</sup> + 1 / (2p<sup>4</sup>)
(iii) Using (a), or otherwise, prove that &alpha;<sup>4</sup> + &beta;<sup>4</sup> &ge; 2 + &radic;2

13. Suppose that J<sub>n</sub> = &int; (0 --> &pi;/2) cos<sup>n</sup>x dx, where n is an integer and n &ge; 1
(i) Prove that J<sub>n</sub> = [(n - 1) / n] * J<sub>n-2</sub>, for n &ge; 2
(ii) By evaluating J<sub>2n</sub> and J<sub>2n+1</sub>, show that:
J<sub>2n+1</sub> / J<sub>2n</sub> = [(2<sup>2</sup> * 4<sup>2</sup> * 6<sup>2</sup> * ... * (2n)<sup>2</sup>) / (3<sup>2</sup> * 5<sup>2</sup> * 7<sup>2</sup> * ... * (2n - 1)<sup>2</sup> * (2n + 1)<sup>2</sup>)] * (2 / &pi;)
(iii) Defining R as J<sub>2n</sub> / J<sub>2n+1</sub>, show that:
&pi; / 2 = [2<sup>2</sup> / (1 * 3)] * [4<sup>2</sup> / (3 * 5)] * [6<sup>2</sup> / (5 * 7)] * ... * [(2n)<sup>2</sup> / ((2n -1) * (2n + 1))] * R
(iv) Remembering that 0 &le; x &le; &pi; / 2, show that 1 &le; R &le; J<sub>2n-1</sub> / J<sub>2n+1</sub>
(v) Hence, show that lim<sub>n--->&infin;</sub>R = 1, and thus derive Wallis' Product for &pi; / 2, which is:
&pi; / 2 = lim<sub>n--->&infin;</sub>[2<sup>2</sup> / (1 * 3)] * [4<sup>2</sup> / (3 * 5)] * [6<sup>2</sup> / (5 * 7)] * ... * [(2n)<sup>2</sup> / ((2n -1) * (2n + 1))]
(vi) Prove also that lim<sub>n--->&infin;</sub> 2<sup>2n</sup>(n!)<sup>2</sup> / [&radic;n * (2n)!] = &radic;&pi;

14. ABC is a triangle in which the ratio of the sides AB:AC is 2:3. The abgle opposite the side AC is 45° larger than the angle opposite the side AB. Show that one of the angles of the triangle is tan<sup>-1</sup>[(3&radic;2 + 2) / 7]

15. By first proving that (1 + itan&theta;)<sup>n</sup> + (1 - itan&theta;)<sup>n</sup> = 2sec<sup>n</sup>&theta;cos n&theta;, show that Re[(1 + itan(&pi; / 8))<sup>8</sup>] = 64(17 - 12&radic;2)

16. Show that the set S = {5, 9, 13, ...} is closed under multiplication.

17. Consider the integral I = &int; x<sup>3</sup> / (x<sup>2</sup> + 1)<sup>3</sup> dx
(a) Find I using the substitution u = x<sup>2</sup> + 1
(b) Find I using the substitution x = tan &theta;
(c) Explain any difference between your results

18. An object is moving in simple harmonic motion about its mean position. At then ends of three successive seconds, the distances from the mean position of the motion to the object are 1, 5 anf 5 metres, each measured in the same direction. Show that the motion has a period of 2&pi; / cos<sup>-1</sup>(3 / 5) seconds, and find the amplitude of motion.

19. Define I<sub>n</sub> = &int; (0 --> &pi;) sin<sup>n</sup>&theta; d&theta; and J<sub>n</sub> = &int; (0 --> &pi;) &int; (0 --> &pi;) sin<sup>n</sup>&theta; d&theta;sin<sup>n</sup>&theta; d&theta;, both for integres n &ge; 1. Prove that, for integers n &ge; 3, J<sub>n</sub> = [&pi;(n - 1) / 2n]I<sub>n-2</sub>, and hence show that J<sub>8</sub> = 35&pi;<sup>2</sup> / 256

20. ABC is an equilateral triangle, and all three vertices lies on the same circle. X is a point on the minor arc BC and AX and BC meet at D.
(a) Show that &Delta;BDX ||| &Delta;ACX
(b) XA = XB + XC
 
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Xayma

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CM_Tutor said:
7. Two circles, C<sub>1</sub> and C<sub>2</sub>, intersect at points K and L. KM is a tangent to circle C<sub>2</sub> and KN is a tangent to circle C<sub>1</sub>, where M lies on C<sub>2</sub> and N lies on C<sub>1</sub>. NL produced meets KM at P and circle C<sub>2</sub> at Q.
(a) Show that KN || MQ
(b) Show that &Delta;KMQ is isosceles
Shouldnt that be , where M lies on C<sub>1</sub> and N lies on C<sub>2</sub> because they are a tangent currently M=N=K?
 

CM_Tutor

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Xayma said:
Shouldnt that be , where M lies on C<sub>1</sub> and N lies on C<sub>2</sub> because they are a tangent currently M=N=K?
Yes, it should - it makes no sense the way I had it (Oddly enough, that's the way it was in the source). Anyway, edited and fixed in original post. Thanks.
 

:: ck ::

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yay cm! ur the best :)

have fun on ur trip ~!
 

CM_Tutor

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Have people tried these questions? Does anyone want to post up solutions, or ask about any of these?
 

CrashOveride

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Can they serioulsy put that LATEX thing on here? For writing up longer stuff its quite frustratign using the currently available functions
 

BillyMak

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7. Two circles, C<sub>1</sub> and C<sub>2</sub>, intersect at points K and L. KM is a tangent to circle C<sub>2</sub> and KN is a tangent to circle C<sub>1</sub>, where M lies on C<sub>1</sub> and N lies on C<sub>2</sub>. NL produced meets KM at P and circle C<sub>2</sub> at Q.
(a) Show that KN || MQ
(b) Show that &Delta;KMQ is isosceles

I found another little boo boo in this question.
Probably a little late, but hey, I'm new to this place.

Surely "NL produced meets KM at P and circle C<sub>2</sub> at Q" should read "NL produced meets KM at P and circle C<sub>1</sub> at Q" ?
 

SeDaTeD

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CM_Tutor said:
13.
(ii) By evaluating J<sub>2n</sub> and J<sub>2n+1</sub>, show that:
J<sub>2n+1</sub> / J<sub>2n</sub> = [(2<sup>2</sup> * 4<sup>2</sup> * 6<sup>2</sup> * ... * (2n)<sup>2</sup>) / (3<sup>2</sup> * 5<sup>2</sup> * 7<sup>2</sup> * ... * (2n - 1)<sup>2</sup> * (2n + 1)<sup>2</sup>)] * (2 / &pi;)
The (2n + 1) at the end there should not have a power of 2.
 

Stefano

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If there arent any solutions they aren't much use.

But thanks anyways Mr. Bin Laden. I mean Osama!
 

FinalFantasy

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Stefano said:
If there arent any solutions they aren't much use.

But thanks anyways Mr. Bin Laden. I mean Osama!
You can just post any questions that you may have problems with in the forum...
 

rabdog

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How do you know if you have done the question correctly or not?
 

rabdog

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I have a problem with question 5 of that first link
Show that Im(z) = (z - z_) / 2
z_ is the conjugate of z

i end up getting Im(z) = yi when it should = y whats going on!?
 

FinalFantasy

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rabdog said:
I have a problem with question 5 of that first link
Show that Im(z) = (z - z_) / 2
z_ is the conjugate of z

i end up getting Im(z) = yi when it should = y whats going on!?
Im(z)=iy
so (z - z_) / 2 is imaginary
u've proven it
 

rabdog

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Im pretty sure that the imaginary part of z is the y part, not including i
 

FinalFantasy

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To prove: Im (z)=(z-z(bar) )\2
let z=x+iy

(z-z(bar))\2=(x+iy-(x-iy) )\2
=iy

hmmm... do u tink the q. mite have a mistake?
like it should be prove Im (z)=(z-z(bar))\2i ?
 

Roobs

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Re: More Questions

Does anyone have a solution to q1 part b...im stuck
 

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