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Addition formulae
(higher level courseware)
When trying to determine the missing data of planar figures we experienced that the trigonometric functions can be used as effective tools. Their general definitions can be given with the help of the unit vector. (Figure 35)
\latex{\overrightarrow{e}=\cos{\alpha}\times}\latex{\text{i}}\latex{+\sin{\alpha}\times}\latex{\text{j}},
where \latex{\overrightarrow{e}} is a unit vector with an azimuth angle of \latex{\alpha}.
The vector operations we became familiar with and their characteristics might provide new identities.
\latex{\sin{\alpha}}
\latex{\overrightarrow e}
\latex{\overrightarrow e(\cos \alpha; \sin \alpha)}
\latex{\left|\overrightarrow e\right|=1}
\latex{\cos{\alpha}}
\latex{\alpha}
\latex{ x }
\latex{ y }
Figure 35
Let us consider the unit vectors \latex{\overrightarrow{e_1}} and \latex{\overrightarrow{e_2}}; let their azimuth angles be denoted by \latex{\alpha} and \latex{\beta}. The angle included between the two vectors is \latex{\alpha – \beta} according to the figure. (Figure 36)
Let us examine what we get if we calculate the dot product of these two vectors. Let us first write down this product based on the definition:
\latex{\overrightarrow{e_1}\times\overrightarrow{e_2}=1\times1\times\cos{(\alpha-\beta)}}.
If we perform this same operation with the help of the coordinates:
\latex{\overrightarrow{e_1}\times\overrightarrow{e_2}=\cos{\alpha}\times\cos{\beta}+\sin{\alpha}\times\sin{\beta}}.
Obviously the two results need to be identical. Based on this the following theorem can be formulated:
\latex{\overrightarrow e}
\latex{\overrightarrow{e_1}(\cos \alpha; \sin \alpha)}
\latex{\alpha}
\latex{\beta}
\latex{\overrightarrow{e_2}(\cos \beta; \sin \beta)}
\latex{ x }
\latex{ y }
Figure 36
THEOREM: The cosine of the difference of two angles can be obtained by adding the product of the cosines of the two angles and the product of the sines of the two angles:
\latex{\cos{(\alpha-\beta)}=\cos{\alpha}\times\cos{\beta}+\sin{\alpha}\times\sin{\beta}}.
\latex{\cos{(\alpha-\beta)}=\\=\cos{\alpha}\times\cos{\beta}=\sin{\alpha}\times\sin{\beta}}
Example 1
Let us give the value of \latex{\cos 15^{\circ}}.
Solution
Let us use that \latex{45^{\circ}-30^{\circ}=15^{\circ},} therefore:
\latex{\cos{15^{\circ}}=\cos{(45^{\circ}-30^{\circ})}=\cos{45^{\circ}}\times\cos{30^{\circ}}+\sin{45^{\circ}}\times\sin{30^{\circ}}=\\=\frac{\sqrt2}{2}\times\frac{\sqrt3}{2}+\frac{\sqrt2}{2}\times\frac{1}{2}=\frac{\sqrt6+\sqrt2}{4}.}
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Based on the addition formula obtained for \latex{\cos(\alpha – \beta )} – it is also called an addition theorem – it is possible to prove further addition theorems.
THEOREM: \latex{\cos{(\alpha+\beta)}=\cos{\alpha}\times\cos{\beta}-\sin{\alpha}\times\sin{\beta}}.
\latex{\cos{(\alpha+\beta)}=\\=\cos{\alpha}\times\cos{\beta}-\sin{\alpha}\times\sin{\beta}}
Proof
Since \latex{\beta} was an arbitrary angle, it can be replaced by \latex{-\beta}.
Then:
Then:
\latex{\cos{(\alpha+\beta)}=\cos{[\alpha-(-\beta)]}=\cos{\alpha}\times\cos{(-\beta)}+\sin{\alpha}\times\sin(-\beta)}.
We knew that \latex{\cos{(-\beta)}=\cos{\beta}} and \latex{\sin{(-\beta)}=-\sin{\beta}}, therefore\latex{\cos{(\alpha+\beta)}=\cos{\alpha}\times\cos{\beta}-\sin{\alpha}\times\sin{\beta}}.
THEOREM: \latex{\sin{(\alpha-\beta)}=\sin{\alpha}\times\cos{\beta}-\cos{\alpha}\times\sin{\beta}}.
\latex{\sin{(\alpha-\beta)}=\\=\sin{\alpha}\times\cos{\beta}-\cos{\alpha}\times\sin{\beta}}
Proof
Let us now use the identity relating to the complementary angles, which tells us that for an arbitrary angle \latex{\varphi} that \latex{\sin \varphi = \cos{(90^{\circ} – \varphi)}}, and let us also use the statement of the previous theorem.
\latex{\sin{(\alpha-\beta)}=\cos{\left[90^{\circ}-(\alpha-\beta)\right]}=\cos{\left[(90^{\circ}-\alpha)+\beta\right]}=\\=\cos{(90^{\circ}-\alpha)}\times\cos{\beta}-\sin{(90^{\circ}-\alpha)}\sin{\beta}=\sin{\alpha}\times\cos{\beta}-\cos{\alpha}\times\sin{\beta}.}
THEOREM: \latex{\sin{(\alpha+\beta)}=\sin{\alpha}\times\cos{\beta}+\cos{\alpha}\times\sin{\beta}}.
\latex{\sin{(\alpha+\beta)}=\\=\sin{\alpha}\times\cos{\beta}+\cos{\alpha}\times\sin{\beta}}
Proof
Since the angle \latex{\beta} in the previous theorem can be replaced by \latex{-\beta}:
\latex{\sin{(\alpha+\beta)}=\sin{\left[\alpha-(-\beta)\right]}=\sin{\alpha}\times\cos{(-\beta)}-\cos{\alpha}\times\sin{-\beta}=\\=\sin{\alpha}\times\cos{\beta}+\cos{\alpha}\times\sin{\beta}.}
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Example 2
Let us verify the following identities:
- \latex{\cos{x+y}+\cos{x-y}=2\times\cos x\times\cos y};
- \latex{\cos{x+y}-\cos{x-y}=-2\times\sin x\times\sin y}.
Solution
Let us use the addition theorems verified for the cosine trigonometric function. According to these:
\latex{\cos{(x+y)}=\cos{x}\times\cos y-\sin x\times\sin y};
\latex{\cos{(x-y)}=\cos x\times\cos y+\sin x\times\sin y}.
\latex{\cos{(x-y)}=\cos x\times\cos y+\sin x\times\sin y}.
By adding the two relations the statement of part a), and by subtracting them the statement of part b) is obtained.
The result of the example makes it possible to factorise the sum or the difference of the cosines of angles. Let us denote \latex{(x+ y)} and \latex{(x – y)} by a single variable:
\latex{x+y=\alpha;\;\;x-y=\beta}.
By adding and subtracting the two equations the angles \latex{x} and \latex{ y } can be expressed in terms of \latex{\alpha} and \latex{\beta}:
\latex{x=\frac{\alpha+\beta}{2};\;\;y=\frac{\alpha-\beta}{2}.}
When taking these into account the following two identities are obtained:
\latex{\cos{\alpha}+\cos{\beta}=2\times\cos{\frac{\alpha+\beta}{2}}\times\cos{\frac{\alpha-\beta}{2}}};
\latex{\cos{\alpha}-\cos{\beta}=-2\times\sin{\frac{\alpha+\beta}{2}}\times\sin{\frac{\alpha-\beta}{2}}.}
\latex{\cos{\alpha}-\cos{\beta}=-2\times\sin{\frac{\alpha+\beta}{2}}\times\sin{\frac{\alpha-\beta}{2}}.}
Exercises
{{exercise_number}}. Give the value of the sine and the cosine of the following angles without using a calculator or a function value table:
- \latex{75^{\circ}};
- \latex{-15^{\circ}};
- \latex{105^{\circ}}.
{{exercise_number}}. Give the angle of inclination of the vectors if the following is given:
- \latex{2\times\sin{\left(\frac{\pi}{4}-\alpha\right)}\times\cos{\left(\frac{\pi}{4}+\alpha\right)}=1-2\times\sin{\alpha}\times\cos{\alpha}};
- \latex{\sin{\alpha}+\sin{\left(\alpha+\frac{2\pi}{3}\right)}+\sin{\left(\alpha+\frac{4\pi}{3}\right)}=0}.
{{exercise_number}}. Verify the following identities:
- \latex{\sin{(x+y)}+\sin{(x-y)}=2\times\sin x\times\cos y};
- \latex{\sin{(x+y)}-\sin{(x-y)}=2\times\cos x\times\sin y}.
{{exercise_number}}. By using the result of the previous exercise verify that:
- \latex{\sin{\alpha}+\sin{\beta}=2\times\sin{\frac{\alpha+\beta}{2}}\times\cos{\frac{\alpha-\beta}{2}}};
- \latex{\sin{\alpha}-\sin{\beta}=2\times\cos{\frac{\alpha+\beta}{2}}\times\sin{\frac{\alpha-\beta}{2}}}.
{{exercise_number}}. Verify that the following holds for the angles \latex{\alpha}, \latex{\beta} and \latex{\gamma} of a triangle:
- \latex{\sin{\gamma}=\sin{(\alpha+\beta)}};
- \latex{\sin{\alpha}\times\sin{\gamma}-\cos{\alpha}\times\cos{\gamma}=\cos{\beta}}.
{{exercise_number}}. Express the following expressions in terms of the trigonometric functions of \latex{\alpha}, \latex{\beta}, \latex{\gamma} :
- \latex{\sin{(\alpha+\beta+\gamma)}};
- \latex{\sin{(\alpha+\beta-\gamma)}};
- \latex{\cos{(\alpha-\beta+\gamma)}};
{{exercise_number}}. The plan of a room is a right-angled triangle. Its legs are \latex{ 3\,m } and \latex{ 4\, m } long. Where shall I stand in order to see each side of the room at the same angle?
