Monday, August 17, 2015

Prove that $\frac{cot(x)}{1-tan(x)} + \frac{tan(x)}{1 - cot(x)}=1 + tan(x) + cot(x)$

Q. Prove: $\frac{cot(x)}{1-tan(x)} + \frac{tan(x)}{1 - cot(x)}=1 + tan(x) + cot(x)$}

A. $$\begin{array}{lll}\\
\frac{cot(x)}{1-tan(x)} + \frac{tan(x)}{1 - cot(x)} &= \frac{\frac{1}{tan(x)}}{1 - tan(x)} + \frac{tan(x)}{1 - \frac{1}{tan(x)}}&\mbox{Expressing } cot(x)=1/tan(x)\\
&= \frac{1}{tan(x)(1-tan(x))} + \frac{\tan^2(x)}{tan(x) - 1}& \mbox{Simplifying fractions.}\\
&= \frac{1}{tan(x)(1-tan(x))} - \frac{\tan^2(x)}{1- tan(x)}& \mbox{Last term as a subtraction }\\
&= \frac{1 - tan^3(x)}{tan(x)(1-tan(x)}& \mbox{Rewriting as a single fraction}\\
&= \frac{(1 - tan(x)){(1 + tan(x)} + \tan^2(x)}{tan(x)(1-tan(x))}& \mbox{Factoring a difference of cubes}\\
&= \frac{(1 + tan(x) +\tan^2(x)}{tan(x)}&\mbox{Cancellation by division}\\
&= cot(x) + 1 + tan(x)& \mbox{Division}\\
&= 1 + tan(x) + cot(x)& \mbox{Rearranging terms.}\\

Derivation of the formula for cos(alpha - beta)

Derivation of the length of a chord

Let an arc on the unit circle with starting point at $ P_0 = (1, 0)$ subtend an angle of $\beta$ radians. Recall that its endpoint will be
$P_1 = (\cos(\beta), \sin(\beta))$. The length of the chord connecting the point (1,0) to the terminal point can be determined using the
distance formula $$d = \sqrt{(\Delta x)^2 + (\Delta y)^2}=\sqrt{(\cos(\beta) - 1)^2 + (\sin(\beta) - 0) ^2}$$. Simplifying we will have
$d = \sqrt{\cos^2(\beta) + \sin^2 (\beta) - 2\cos(\beta) + 1}= \sqrt{2- 2\cos(\beta)}.$

The fundamental formula for the cosine of the difference in angles: $\cos(\alpha - \beta)$

Now let $\alpha$ be an angle greater than $\beta$. The new endpoint will be at $P_2 =(\cos(\alpha), sin(\alpha))$. Consider the chord $P_1P_2$.
By the distance formula, the length of this chord will be given by $\sqrt{(\cos (\alpha) - \cos(\beta))^2 +(\sin(\alpha)-\sin(\beta))^2}$
and expanding we obtain
$\sqrt{(\cos^2(\alpha) - 2 cos(\alpha)cos(\beta) + cos^2(\beta) + (\sin^2(\alpha) - 2 sin(\beta)\sin(\alpha)+\sin^2(\beta))} $

Further simplifying we obtain $\sqrt{(\cos^2(\alpha) + \sin^2(\alpha) + \cos^2(\beta) + \sin^2(\beta) - 2\cos(\alpha) cos\beta - \sin(\alpha) \sin(\beta)}$.
and finally we have $d = \sqrt{2 -2 (\cos(\alpha) cos\beta + \sin(\alpha) \sin(\beta)}$
But this chord lengthis also given by $\sqrt{ 2 - 2 cos(\alpha - \beta)}$.

Equating the two expressions , we obtain the formula

$$\cos(\alpha - \beta) = \cos(\alpha) cos(\beta) + \sin(\alpha) \sin(\beta)$$

This is an important result. Let us try to derive further formulas from this.
Substiture $-\beta$ for $\beta$. We sould then have $\cos(\alpha - (-\beta)) = \cos(\alpha) cos(-\beta) + \sin(\alpha) \sin(-\beta)$.
or $$\cos(\alpha + \beta)) = \cos(\alpha) cos(\beta) - \sin(\alpha) \sin(\beta)$$ using the facts that the cosine function is even $\cos(-alpha) = \cos(alpha)$
that the sine function is odd $sin(-\beta) = -sin(beta)$.We can combine the two identities as $$\cos(\alpha \pm \beta) = cos(\alpha) cos(\beta) \mp \sin(\beta) sin(\alpha)$$.

Wednesday, August 12, 2015

The angle between the hour and minute hands of a clock.

Q. The time shown in the clock is 7:35 what is the angle between the hour and minute hands of a clock?

A. Consider the simpler problem if the time is 7:00. As there are 12 hours in a complete revolution of the hour hand, one hour subtends an angle
of 360/12 = 30 degrees. Thus, if H is the number of hours, the angle made by the hour hand from the noon positioon would be 30H.

Now let us consider the minute hand. if M is the number of minutes, consider that if M is 15 minutes , the angle which the minute hand makes is
90 degrees. This means that each minute ssubends an angle of 90/15 = 6 degrees or 6M.

But each revolution of the minute hand imparts 1/12 revolution of the hour hand. In other words, in the time covered by the minute hand, the hour hand will cover an
addittional 6M/12 or M/2 degrees.

The angle between the hour hand and minute hand for a time reading of H:M will then be |30H + M/2 - 6M|. If this angle is greater than 180 degrees,
we take the difference from 360 degrees.

Getting back to our original problem, for a time of 7:35, the angle beween the hour and minute hands is |30(7) + 35/2 - 6(35)| = 17.5 degrees.

Monday, August 20, 2012

Statistics: basic descriptive statistics, mean , variance, summation

Statistics Problem Set Aug-21-2012

1. Which of the following formulas measure symmetry of a sample data distribution?

(a)$(1/n) \sum (x-\overline{x})^2$ (b) $ (1/n) \sum (x-\overline{x})^3$ (c)$ (1/n) \sum (x-\overline{x})^4$ (d.) Not listed

2. The following were determined for a sample data: n = 10, min=-2, max= 10, sd = 3,
$\overline{x}=5$. The data is invalid since

$max > \overline{x} + (n-1) sd.$ (b) $min < \overline{x} - (n-1) sd$ (c) median is not specified (d) None of the above.

3. A sample has current mean mean 4.0 with sample size 10. If 5 is added to all sample values, the recomputed mean will :

stay the same! (b) will increase by 5.0 (c) will decrease by 5 (d)will increase by 0.5 (e) None of the above

4.A sample has current variance of 2.0 with sample size 10. If 5 is added to all sample values, the recomputed variance will :

stay the same! (b) will increase by 5.0 (c) will decrease by 5 (d)will increase by 0.5 (d) None of the above.

5. The formula $e^{[(1/n)\sum_{i=1}^{i = n} \ln(x_i)}$ is another way of solving for the

(a) HM (b) GM (c) AM (d) RMS.

6. A new data value 2.5 is added to a sample with mean 3 and sample size 10. The recomputed mean will

increase (b) decrease (c) stay the same (d) None of the above

7. A sample has a sample size 10 and median 50, with all sample values unique. If the minimum value in the sample is removed,
the new median will

stay the same! (b) increase (c) decrease (d) None of the above.

8. Express (-1) + 2 + (-3) + 4 + (-5) + 6 in summation form with index k $ of summation varying from 1 to 6. You cannot use
a variable X as these are not stored in an a vector or array!.

9. If X = c( 4,5,6,7,8), what is the value of $\sum_{i=1}^4 (x_i)(x_{i-1})$?.

10.If X is the same above and Y= c(2,3,1,0,4) what is the value of $\sum_{i=1}^5 {(x_i -\overline{x}) (y_i-\overline{y})}$?

Solutions next week! Enjoy first solving.

Tuesday, July 31, 2012

Descriptive Statistics

Modified True or False. If the statement is true, write T and if the statement if false , write F and explain why!

Q1. The values computed for a sample, arithmetic mean =-4.5, and root mean square= -4 are valid since the root mean square is always greater than or equal to the arithmetic mean.

Q2. Variances computed from a sample were sample variance = 0.4629, and population variance = 0.5141 are valid.

Q3. The variances computed with a sample size n = 10, population variance = 0.4629, and sample variance = 0.60, are valid.

Q4. Various means computed for a sample: harmonic mean, HM = 0.2, geometric mean, GM = 0.2, arithmetic mean, AM = .3,
root mean square, RMS = .4 are valid since they satisfy \(HM\le GM\le AM\le RMS \).

Q5. If the variance of a sample X is 5.3 then the variance of \(Y = -2 (X + 5)\) is \(2 \cdot 5.3 = 10.6\).


1. F. The reasoning may sound valid but the root mean square is never negative!

2. F. Sample variance is always greater than population variance!

3. F. Sample varince may be obtained using the formula: $$\sigma_{n-1}^2= \frac{n}{n-1} \sigma_n^2$$. But $$\frac{10}{9} 0.4629 = 0.5141$$ which is not equal to 0.60.

4. Although the given values does satisfy the inequality for various means, whenever any two of the means are equal, then all means should be equal!

5. If X has the variance \(V_X\), then \(Y = a(X + b)\) will have variance $$a^2 V_X$$. Therefore, the varice of Y will be 4 times the variance of X or 4 (5.3)= 21.2.

Monday, May 7, 2012

Problems with complex numbers.

Q1. Convert the complex number z = 6 + 8j to polar form.

A1. The magnitude \(|z|= \sqrt{6^2 + 8^2}= \sqrt{36+ 64} = \sqrt{100} = 10.\)

Since both x and y components are positive, the complex number is in the first quadrant.
The argument \(\theta = atan(\frac{8}{6}) = 0.9273\), in radians. Multiply by \( 180/\pi \) to obtain the argument degrees. The value would be \(53.3^\circ\).

Q2, Convert to rectangular form the complex number in polar form \( 12 \angle 210^\circ \).

A2. The complex number is in the third quadrant. This means that both components are negative.
\(x = 12 \cos(210^\circ) = -10.392, y = 12 \sin(210^\circ) = -6.0\),from which \(z = -10.392 - 6j.\)

Saturday, May 5, 2012

Fomula mass of chemical compound

Q. Calculate the formula mass of the compounds KBr and \( NH_4CN\)

A. The atomic masses of K and Br are respectively 39.102 and 79.904 respectively. Thus the formula mass is 39.102+79.904= 119.006 amu.

On the other hand, NH4CN has a formula mass calculated in the following table:

Component AmuTotal
N 14.0067114.0067
C 1212
N 14.006714.0067

The formula mass is then 44.107.