Some Mathematical Tips for the New Year

Happy New Year everybody! At this time last year I suggested some new year resolutions to help you in your studies. They are of course valid at all times, and if you didn’t see them, then have a look now.

This time I’ve got some tips which you might want to note down; they cover specific areas of the syllabus, although they aren’t applicable to all of HL, SL, and Studies.

Degrees or Radians?

Well, it depends on what the question is asking you to do. Generally, when dealing with plane geometry (let’s say, angles in triangles or circle theorems), I don’t think anyone envisages actual angles measured in radians, and so they use degrees. The main exception is the use of the two formulae that give the arc length and area of a sector: arc length = , area = {1 \over 2}r2θ, where θ must be in radians. The reason for using radians is that the formulae are more complicated when the angle is expressed in degrees. Of course, when differentiating and integrating trigonometric functions you must use radians.

Differentiation and integration of sin and cos

It’s easy to remember that, whether integrating or differentiating, sin becomes cos and cos becomes sin—but somewhere in there are some negatives. How can you remember which is which? Try a simple mnemonic such as ‘Some Individuals Can’t Digest Nuts.’ This leads to ‘Sin Integrated, Cos Differentiated gives Negative.’

And while we’re on trigonometric functions, you should be able to reproduce the following table of exact values:






A neat little way of memorising the first row is that it forms the sequence:

{{\sqrt 0 } \over 2}{{\sqrt 1 } \over 2}{{\sqrt 2 } \over 2}{{\sqrt 3 } \over 2}{{\sqrt 4 } \over 2}

And then the second row, with the values of cos, is the exact reverse of the sin row. The tan row can be quickly calculated by dividing the sin values by the cos values (because \tan \theta  = {{\sin \theta } \over {\cos \theta }}).

Finally, how about those…

Formulae for Points on a Straight Line

…. which all look similar. The main ones you need to know, given two points ({x_1},{y_1}){\rm{ and }}({x_2},{y_2}), are:

The midpoint is \left( {{{{x_1} + {x_2}} \over 2},{{{y_1} + {y_2}} \over 2}} \right)

The gradient of the line joining the points is {{{y_2} - {y_1}} \over {{x_2} - {x_1}}}

The distance between the points is \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}

The equation of the line through the points is y - {y_1} = m(x - {x_1}), where m is the gradient.

But the first three can be easily seen when looking at a diagram where I have plotted two points and the line between them:







  • To find the midpoint, simply work out the number halfway between the x coordinates (i.e., the mean value), and then the same for y. Halfway between 2 and 6 is 4, halfway between 4 and 11 is 7.5, and so the midpoint is (4, 7.5). This works just the same if any of the coordinates are negative.
  • To find the gradient, use the old tip of ‘rise over run’. y has increased from 4 to 11, which is 7, and x from 2 to 6, which is 4. So the gradient is
    {7 \over 4} = 1.75. If the line is ‘going downhill’, don’t forget that the increase in y will be negative, giving a negative gradient.
  • And the distance between the two points is found using Pythagoras’ Theorem. We know the lengths of the two red lines are 4 and 7, so the distance is \sqrt {{4^2} + {7^2}}  = \sqrt {65}

Have you got any useful tips to help you remember formulae, methods, and any other mathematical nasties? Let me know, and I’ll include them in a later blog. In the meantime—have a great year!

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