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1. Trigonometry

The graph of  \cos{ \theta }

It is important to be aware of the different graphs which are associated with the trigonometric functions that one will be using throughout this work. One could, if one was inclined in such a manner, calculate the points for the graphs of the trigonometric functions using a calculator, and then plot the graph; this is, however, inefficient.

One should already be aware of the periodicity of the graphs of the trigonometric functions, more simply put, one should be aware that the graph of  y = \cos{ \theta } has a manner of repetition to it, which enables one to plot the graph (in a sketch) such that one does not need to calculate many of the points.

During the GCSE course, one will have been aware that one can calculate the value of the sine, cosine or tangent of an angle using a right-angled triangle. This can only be used with angles which are between  0^{ \circ } , and  90 ^{ \circ } (not inclusive). THis can be extended now using a unit circle.

Consider the idea (presented in Core 1) of the circle being an infinite set of right-angled triangles, whose vertices represent the centre, a point upon the circumference of the circle, and a point upon a radius. This can be used now (as one might predict) to calculate the sine, cosine, and tangent of angles which are above (or indeed below) the aforementioned bound.

Consider the diagram below:

Figure 001

In the diagram one can observe that there are two right-angled triangles. Consider the triangle whose vertex which is upon the circumference of the circle has coordinates (3, 4). Let the angle which is made by this triangle's hypotenuse, and the x-axis be denoted by  \theta . One can calculate, using the knowledge which one has from GCSE (regarding the definition of cosine), the cosine of this angle:

 \cos{ \theta } = \frac{Adjacent}{Hypotenuse} = \frac{3}{5} = 0.6

(Note that this triangle [and the circle in the diagram] are not using the idea of "unit radius" or "unit hypotenuse", but the approach is still valid).

This is an example of the calculation of the cosine of an angle whose angle lies within the bounds which one has seen previously. What about the other right-angled triangle?

This represents a different angle. If one is to consider the angles which all the possible right-angled triangles can make with the x-axis, one would count in infinitesimally small measures of rotation beginning at the x-axis, and working around in an anti-clockwise manner. This would suggest that the angle which is represented by the second triangle is indeed above the upper bound for the set of angles which the right-angled triangle rule would work for.

There is a triangle however, and therefore one could calculate the cosine (and indeed if one was to input an angle which was greater than the upper bound which was mentioned previously into a calculator, and then apply the "cosine" function, one would obtain a value).

Consider the idea of a negative angle. If one is to use the same idea of counting an angle from the x-axis, but in the clockwise direction, it is true that one will obtain a negative angle.

The triangle is now accessible. Let the angle between the x-axis, and the hypotenuse of the triangle (in a clockwise manner) be denoted as  \phi . Now one can say that the angle is equal in the context of magnitude to  -180^{ \circ } - \phi .

One can predict the sign of the cosine of any angle using the idea of a circle. Consider that the adjacent side of the angle (that has been previously considered) (on the Cartesian plane) will be of negative value when it is on the left-hand-side of the y-axis. This means that there are two quadrants in which there will be a negative value for the adjacent side. In a circle of unit radius, the cosine is equal to th adjacent side, as:

 \cos{ \alpha } = \frac{Adjacent}{Hypotenuse} = \frac{Adjacent}{Radius} = \frac{Adjacent}{1} = Adjacent

This suggests that the angles which are represented by right-angled triangles in these two quadrants (left of the y-axis) will have negative cosine values.

One must also consider the idea of a cosine of an angle which is greater than  360^{ \circ } . It is simple using the definition discussed before. One can merely follow an angle around the circle, and then make the right-angled triangle as usual, however one can calculate the angle which this hypotenuse makes with the x-axis (this will be small enough such that one can use the aforementioned techniques to determine the size, etc. of the cosine of this angle.

One can now consider the graph of  y = \cos{ x } .

Figure 002

One can note that the function never exceeds the value of 1, and is never below -1. The function is oscillating between values, one a predictable path, and is said to be periodic. Now consider the idea of how long it takes to repeat the pattern. This is the same idea of a wavelength, one can see that from a "peak" to a "peak" is  360^{ \circ } , and therefore one can say that the graph of  y = \cos{ x } has period  360^{ \circ } .

This leads to the periodic property of the cosine function. As the graph repeats every  360^{ \circ } , one can therefore say that:

 \cos{ \theta } \equiv \cos{ \left( 360^{ \circ } - \theta \right) } .

Example

1. Find the maximum, and minimum values, and also the period of the following function:

 y = \cos{ 2x } + 4 .

One can reflect upon the idea of the transformations of graphs in order to do this question.

Remember that:

 f(x) \to f(2x)

Corresponds to a stretch, parallel to the x-axis, scale factor  \frac{1}{2} . Now one can say that there are 2 of the "wavelengths" during a period of  360^{ \circ } . Hence the period of the function is:  \frac{ 360^{ \circ }}{2} = 180^{ \circ } .

Now one can consider the following (for the calculation of the minimum and maximum values):

 f(x) \to f(x) + 4

Corresponds to a translation of "4" in the positive y-direction. Hence one can predict that (as there is no effect on the minimum and maximum values of the curve by the stretch parallel to the x-axis) the minimum value is -1 + 4 = 3, and the maximum is 1 + 4 = 5.

One must be vigilant of the possible trick of a stretch parallel to the y-axis, which would change the minimum and maximum values.

The graphs of  \sin{ \theta } and  \tan{ \theta }

One can use the same idea as previously used with the cosine to calculate the sine, and the tangent. It is easy now to consider the places where the sine will be negative. Consider:

 \sin{ \alpha } = \frac{Opposite}{Hypotenuse}

As the hypotenuse of the triangle is equal to 1, the sine of an angle is the opposite side (assuming that one is using the circle of unit radius).

The opposite side (to the angle at the centre of the circle) is negative when the triangle is below the x-axis, and therefore one can predict those angles for which the sine is negative.

As the tangent relies of both the opposite and the adjacent:

 \tan{ \alpha } = \frac{Opposite}{Adjacent}

One can consider that in the two quadrants in which only one of the variable sides is negative will angles have negative tangents.

Below are the graphs of  y = \sin{x} , and  y = \tan{x} .

Figure 003


Figure 004

One can see that as with the cosine function, the sine function, and the tangent function are periodic. The sine function has period  360^{ \circ } , and the tangent function has period  180^{ \circ } .

Example

1. Calculate the minimum, and maximum values, along with the period of the following functions:

a)  y = \frac{1}{2 \sin{x} + 3}

b)  y = tan{x} .

a) For the period of this function one can ignore the fractional part, hence one must find the period of the denominator.

 f(x) \to 2 \times f(x)

Corresponds to a stretch, parallel to the y-axis, scale factor 2.

 f(x) \to f(x) + 3

Corresponds to a translation of 3 in the positive y-direction.

Neither of these will effect the period of the function, and therefore one can state that the function has period  360^{ \circ } .

Now for the minimum and maximum. This question is somewhat of a trick (one might say). If one is to divide 1 by a very tiny positive value, one will get a large positive value, and with negative values it is the same (magintudinally speaking), therefore there are no values for the minimum and maximum.

b) The tangent function does not have a value for minimum or maximum (one can see this from the graph). The period is  180^{ \circ} .

Exact values of some trigonometric functions

Although the use of a calculator is permitted in the examination, it is useful to be able to quote the sine, cosine and tangent of certain angles in the exact form. One can use two different right-angled triangles in order to calculate the sine, cosine and tangent of the angles:  30^{ \circ } ,  45^{ \circ } , and  60^{ \circ } .

If one was to have a right-angled triangle which was also isosceles, it would have angles of  90^{ \circ } ,  45^{ \circ } , and  45^{ \circ } . Consider a triangle like this which has two sides of 1 unit. The other side will be (by Pythagoras)  \sqrt{1^{2} + 1^{2}} = \sqrt{2} .

This means that one can calculate the sine, cosine, and tangent of  45^{ \circ } . Consider that the opposite side to a  45^{ \circ } angle will always be the sides of length 1, and therefore:

 \sin{45^{ \circ }} = \frac{Opposite}{Hypotenuse} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}

Similarly:

 \cos{45^{ \circ }} = \frac{Adjacent}{Hypotenuse} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}

Tangent can also be calculated:

 \tan{45^{ \circ }} = \frac{Opposite}{Adjacent} = \frac{1}{1} = 1 .

One should remember these values.

The other triangle is constructed in a slightly more complex manner. Construct first an equilateral triangle of side 2 (hence all interior angles are  60^{ \circ } .

Now, drop a line which bisects one angle, and the opposite side, leaving two right-angled triangles whose hypotenuses are 2, and have a side of length 1; also they have angles of  90^{ \circ} ,  60^{ \circ} , and  30^{ \circ } . By Pythagoras, the third side is:  \sqrt{2^{2} - 1^{2}} = \sqrt{4 - 1} =  \sqrt{3} .

Now one is able to calculate the sine, cosine, and tangent:

 \sin{30^{ \circ }} = \frac{Opposite}{Hypotenuse} = \frac{1}{2}

 \cos{30^{ \circ }} = \frac{Adjacent}{Hypotenuse} = \frac{\sqrt{3}}{2}

 \tan{30^{ \circ }} = \frac{Opposite}{Adjacent} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}

Also:

 \sin{60^{ \circ }} = \frac{Opposite}{Hypotenuse} = \frac{\sqrt{3}}{2}

 \cos{60^{ \circ }} = \frac{Adjacent}{Hypotenuse} = \frac{1}{2}

 \tan{60^{ \circ }} = \frac{Opposite}{Adjacent} = \frac{\sqrt{3}}{1} = \sqrt{3}

One should learn the values for the sine, cosine and tangent of the right-angle. It is easy to see the value from their graphs (shown previously):

 \sin{90^{ \circ }} = 1

 \cos{90^{ \circ }} = 0

The tangent of the right-angle is undefined.

Example

1. Find the exact value of the following:

a)  \sin{495^{ \circ }}

b)  \cos{150^{ \circ }}

c)  \tan{240^{ \circ}} .

a) One must first observe that this is above one full rotation, therefore one must subtract multiples of  360^{ \circ } such that this value can be dealt with.

 495^{ \circ } - 360^{ \circ } = 135^{ \circ }

Therefore:

 \sin{495^{ \circ }} = \sin{135^{ \circ }}

If one observes the graph of the sine function, one can see that within each half of a period ( 180^{ \circ} ) where one begins at  0^{ \circ } , the graph is symmetrical about the mid point. This means that within the domain of  0^{ \circ } \to 180^{ \circ } , the sine function is symmetric about  x = 90^{ \circ } . This means that one can state:

 \sin{ \alpha } \equiv \sin{ 180^{ \circ } - \alpha }

(This will be covered in more depth later in these notes).

Hence:

 \sin{135^{ \circ }} = \sin{180^{ \circ } - 135^{ \circ }} = \sin{ 45^{ \circ }} = \frac{\sqrt{2}}{2}

Therefore:

 \sin{495^{ \circ }} = \frac{\sqrt{2}}{2} .

b)  \cos{150^{ \circ }} .

One might consider using a circle of unit radius for this question. If one was to sketch the circle, and then construct a sketch of the right-angled triangle which represents this angle, one would get a triangle with angle  30^{ \circ } at the centre, but whose adjacent side is negative, hence:

 \cos{150^{ \circ }} = - \cos{30^{ \circ }} = - \frac{\sqrt{3}}{2} .

c)  \tan{240^{ \circ }} .

Again, one could use this idea of the circle of unit radius for this question. The right-angled triangle will be of angle  60^{ \circ } at the centre, and will be in one of the quadrants that the tangent is positive, therefore:

 \tan{240^{ \circ }} = \tan{60^{ \circ}} = \sqrt{3} .

Symmetry properties of the graphs of  \cos{ \theta } ,  \sin{ \theta } and  \tan{ \theta }

It is evident from the shapes (periodicity) of the graphs of the sine, cosine and tangent functions that there will be some form of symmetry about the value of the function with reference to the value of the input.

First consider the graph of:

 y = \cos{ \theta }

This is evidently symmetrical about the y-axis, and therefore, by the rules of transformation of graphs, one might say:

 \cos{ \theta } = \cos{ - \theta }

Therefore the cosine function is an even function.

Another idea is to consider the translation of the aforementioned graph in the theta-direction. If one is to translate by  180^{ \circ } in the positive theta-direction, one will have a graph whose positive values are magnitudinally equal to the negative values of the graph of the cosine function, and whose negative values are magnitudinally equal to the corresponding positive values of the cosine function graph, hence:

 \cos{ \left( \theta - 180^{ \circ } \right) } = - \cos{ \theta }

This is called the translation property.

Another property is the supplementary property, which can be derived using the afore-demonstrated proofs.

Consider:

 \cos{ \left( 180^{ \circ } - \theta \right) }

One can rewrite this:

 \cos{ \left( 180^{ \circ } - \theta \right) } = \cos{\left( - \left( \theta - 180^{ \circ } \right) \right) } = \cos{ \left( \theta - 180^{ \circ } \right) } = - \cos{ \theta } .

Due to the similar nature of the sine and cosine functions, they have similar properties.

Consider the following:

 \sin{ \left( - \theta \right) } = - \sin{ \theta }

 \sin{ \left( 180^{ \circ } - \theta \right) } = \sin{ \theta }

 \sin{ \left( \theta \pm 360^{ \circ } \right) } = \sin{ \theta }

 \sin{ \left( \theta - 180^{ \circ } \right) } = - \sin{ \theta } .

There are similar properties for the tangent function:

 \tan{ \left( \theta \pm 180^{ \circ } \right) } = \tan{ \theta }

 \tan{ \left( 180^{ \circ } \right) } = - \tan{ \theta }

 \tan{ \left( - \theta \right) } = - \tan{ \theta }

Solving equations involving trigonometric functions

One must be careful when solving equations involving trigonometric functions as there are an infinite number of solutions (due to the periodic nature of the trigonometric functions).

If one was to solve the equation:

 \cos{x} = \frac{1}{2} ,

it might be that one would simply do:

 \cos^{-1}{\frac{1}{2}} ,

and quote the result. This is partially correct. The calculator will produce the result of  30^{ \circ } , however as one has seen, the values for which  \cos{x} = \frac{1}{2} are the intersections of the line  y = \frac{1}{2} , with the graph of the cosine function. There are infinitely many intersections, and therefore infinitely many results.

One will therefore usually encounter a range for the results (so as not to involve the description of the infinitely many results).

Example

1. Solve the following equations (for  -360^{ \circ } \le x \le 360^{ \circ } ):

 \cos{x} = \frac{1}{2} .

First one should use the calculator (or one's knowledge, in this case) to obtain the "simplest" result:

 \cos^{-1}{\frac{1}{2}} = 30^{ \circ }

Now one can use the properties which were discussed earlier to obtain the other results.

 \cos{ - \theta } = \cos{ \theta }

Therefore:

 x = \pm 30^{ \circ }

Now one can consider the periodic property:

 \cos{ \theta \pm 360^{ \circ }} = \cos{ \theta }

Therefore:

 x = -30^{ \circ } + 360^{ \circ } = 330^{ \circ } ,

and:

 x = 30^{ \circ }  + 360^{ \circ } = -330^{ \circ } .

Therefore:

 x = \pm 30^{ \circ } , \ \pm 330^{ \circ }

2. Solve the following equation (for  -360^{ \circ } \le x \le 360^{ \circ } ):

 \sin{x + 30^{ \circ }} = 1 .

First one should allow a different variable to be the  x + 30^{ \circ } :

 Let \ y = x + 30^{ \circ }

Therefore:

 \sin{y}= 1

Hence:

 \sin^{-1}{1} = y = 90^{ \circ }

Now one can use the properties of the sine function.

Using the periodic property:

 \sin{ \theta \pm 360^{ \circ }} = \sin{ \theta }

Therefore:

 \sin{90^{ \circ }} - 360^ { \circ } = -270^{ \circ }

Now one can say:

 y = 90^{ \circ } , \ -270^{ \circ }

Therefore:

 x = 70^{ \circ }, \ -290^{ \circ }

Always remember, with questions of this nature, to convert the answers back to the original variable, and also to ensure that all answers quoted are within the mentioned bounds.

3. Solve the following equation (for  -360^{ \circ } \le x \le 360^{ \circ } ):

 \tan{2x + 30^{ \circ }} = 1 .

As previously, one can introduce a new variable:

 Let \ y = 2x + 30^{ \circ } .

With equations of this nature (where it is relatively difficult to discern which bounds one should adhere to with the new variable), one can construct new bounds:

 If \ x = \pm 360^{ \circ }

Therefore:

 y = -690^{ \circ } , \ or \ 750^{ \circ } .

Now:

 \tan{y} = 1

Hence:

 \tan^{-1}{1} = y = 45^{ \circ } .

Now one can apply the properties:

 \tan{ \theta \pm 180^{ \circ }} = \tan{ \theta } .

Hence:

 y = 45^{ \circ } + 180^{ \circ } = 225^{ \circ }

 y = 225^{ \circ } + 180^{ \circ } = 405^{ \circ }

 y  = 405^{ \circ } + 180^{ \circ } = 585^{ \circ}

Also:

 y = 45^{ \circ } - 180^{ \circ } = -135^{ \circ }

 -135^{ \circ } - 180^{ \circ } = -315^{ \circ }

 -315^{ \circ }  - 180^{ \circ } = -495^{ \circ }

 -495^{ \circ } - 180^{ \circ } = -675^{ \circ }

Therefore:

 y = -675^{ \circ } , \ -495^{ \circ } , \ -315^{ \circ } , \ -135^{ \circ } , \ 45^{ \circ } , \ 225^{ \circ } , \ 405^{ \circ } , \ 585^{ \circ }

Therefore:

 x = -352.5^{ \circ } , \ -262.5^{ \circ } , \ -172.5^{ \circ } , \ -82.5^{ \circ } , \ 7.5^{ \circ } , \ 87.5^{ \circ } , \ 187.5^{ \circ } , \ 277.5^{ \circ } .

It can be useful to draw a sketch graph such that one can check the number of (and approximate values of) the roots to the equation.

Relations between the trigonometric functions

When one refers to an "identity", one is referring to a statement of "identical equality". This means that if one gives any value (within the specified set) for the unknown(s), the statement is still true; for example:

 3x - x + 4 \equiv 2x + 4 .

Now:

 \sin{ \theta } = \frac{Opposite}{Hypotenuse}

 \cos{ \theta } = \frac{Adjacent}{Hypotenuse}

 \tan{ \theta } = \frac{Opposite}{Adjacent}

One might notice that if one is to do the following, one will obtain the expression for tangent:

 \frac{ \sin{ \theta } }{ \cos{ \theta } } = \frac{ \left( \frac{Opposite}{Hypotenuse} \right) }{ \left( \frac{Adjacent}{Hypotenuse} \right) } = \frac{Opposite}{Adjacent}

Hence:

 \tan{ \theta } \equiv \frac{ \sin{ \theta } }{ \cos{ \theta } } .

Another identity is the Pythagorean identity, as one might expect, this involves Pythagoras' Theorem. Consider the unit circle idea which was raised at the beginning of this section in these notes. It was demonstrated that the two shorter sides of the right-angled triangles are  \sin{ \theta } , and  \cos{ \theta } ; and the hypotenuse is 1 (as it is the radius). Now one can apply Pythagoras' Theorem, giving the identity:

 \cos^{2}{ \theta } + \sin^{2}{ \theta } \equiv 1 .

These identities can be used to simplify other statements, and also to allow for the solution of more complex trigonometric equations.

Example

1. Express the following in terms of  \cos^{2}{ \theta } :

 \tan^{2}{ \theta } .

One can merely use the identities:

 \tan^{2}{ \theta } = \left( \frac{ \sin{ \theta } }{ \cos{ \theta } } \right) ^{2} = \frac{ \sin^{2}{ \theta } }{ \cos^{2}{ \theta } }

 \frac{ \sin^{2}{ \theta } }{ \cos^{2}{ \theta } }  = \frac{ 1 - \cos^{2}{ \theta } }{ \cos^{2}{ \theta } } = \frac{1}{ \cos^{2}{ \theta }} - 1


(The conversion of  sin^{2}{ \theta } , to  1 - \cos^{2}{ \theta } was merely a re-arrangement of the Pythagorean identity).


2. Solve the following equation (for  -180^{ \circ } \le x \le 180^{ \circ } ):

 2 - 2 \sin{x} = \cos^{2}{x} .

One can re-arrange, and reform to get this into a single trigonometric function (in this case, sine):

 2 - 2 \sin{x} = 1 - \sin^{2}{x}

 1 - 2 \sin{x} + \sin^{2}{x} = 0

 \sin^{2}{x} - 2 \sin{x} + 1 = 0

 Let \ u = \sin{x}

Therefore:

 u^{2} - 2u + 1 = 0

Hence:

 (u - 1)(u - 1) = 0

Hence:

 u = 1

Hence:

 \sin{x} = 1

Hence:

 x = 90^{ \circ }

This is the only root within the boundary. One might also consider using a sketch graph to check the number of roots. Do not try to divide by a trigonometric function (to reduce an equation to a linear trigonometric function) as this would possible cause some roots to be "lost".


Also See

Read these other OCR Core 2 notes:

  1. Trigonometry
  2. Sequences
  3. The binomial theorem
  4. The sine and cosine rules
  5. Integration
  6. Geometric sequences
  7. Exponentials and logarithms
  8. Factors and remainders
  9. Radians
  10. The trapezium rule
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