Vaša košarica je prazna
Applications of the addition formulae
(higher level courseware)
(higher level courseware)
By using the theorems proven in the previous chapter we can proceed even in two directions. On one hand we can examine the relations relating to the value of \latex{\alpha+\beta} and \latex{\alpha-\beta} in the case of the tangent and cotangent trigonometric functions, and on the other hand we have the option to examine the trigonometric functions of \latex{2\alpha}. The following theorems refer to these.
THEOREM: \latex{\tan \left(\alpha +\beta \right) =\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \times \tan \beta }}.
\latex{\tan \left(\alpha +\beta \right) =\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \times \tan \beta }}.
Obviously the theorem holds only for angles for which both sides of the equality have a meaning.
Because of the left-hand-side \latex{\cos \left(\alpha +\beta \right) \neq 0}.
On the right-hand side \latex{ \cos\alpha\neq 0}, \latex{\cos\beta\neq 0 } and \latex{\tan \alpha \times \tan \beta \neq 1}.
Proof
We use the definition of the tangent function in the proof:
\latex{\tan \left(\alpha +\beta \right)=\frac{\sin \left(\alpha +\beta \right)}{\cos \left(\alpha +\beta \right)}=\frac{\sin \alpha \times \cos \beta +\cos \alpha \times \sin \beta }{\cos \alpha \times \cos \beta -\sin \alpha \times \sin \beta }}.
Because of the conditions both the numerator and the denominator of the fraction can be divided by \latex{\cos \alpha \times \cos \beta}.
\latex{\tan \left(\alpha +\beta \right) =\frac{\frac{\sin \alpha \times \cos \beta }{\cos \alpha \times \cos \beta }+\frac{\cos \alpha \times \sin \beta }{\cos \alpha \times \cos \beta }}{\frac{\cos \alpha \times \cos \beta }{\cos \alpha \times \cos \beta }-\frac{\sin \alpha \times \sin \beta }{\cos \alpha \times \cos \beta } } }.
By simplifying the fractions and using the definition of the tangent function the statement to be proven is obtained:
\latex{\tan \left(\alpha +\beta \right) =\frac{\tan\alpha +\tan \beta }{1-\tan \alpha \times \tan \beta } }.
Example 1
Let us give the value of \latex{\tan75°}.
Solution
Since \latex{ 75º = 30º + 45º }, by using the addition theorem and by rationalising the resulting fraction, and then by simplifying:
\latex{\tan 75°=\tan \left(30°+45°\right)=}
\latex{=\frac{\tan 30°+\tan 45°}{1-\tan 30°\times \tan45°}=\frac{\frac{\sqrt{3} }{3}+1 }{1-\frac{\sqrt{3} }{3} }=\frac{\sqrt{3}+3 }{3-\sqrt{3} }=}
\latex{=\frac{\left(\sqrt{3}+3 \right)^2 }{9-3}=\frac{\left(1+\sqrt{3} \right)^2 }{2}=2+\sqrt{3}}.
⯁ ⯁ ⯁
The statement relating to the difference is also obtained similarly.
THEOREM: \latex{\tan \left(\alpha +\beta \right) =\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \times \tan \beta }},
where \latex{\cos \left(\alpha -\beta \right)\neq 0}; \latex{\cos \alpha \neq 0}; \latex{\cos \beta \neq 0}; \latex{\tan \alpha \times \tan \beta \neq -1}.
\latex{\tan \left(\alpha +\beta \right) =\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \times \tan \beta }}.
Proof
Let us write the negative of the angle, i.e.\latex{-\beta} instead of the angle \latex{\beta} everywhere in the formula verified in the previous theorem, and let us use the relation \latex{\tan(-\beta)=-\tan\beta}.
\latex{\tan \left(\alpha +\beta \right) =\frac{\tan \alpha +\tan(- \beta) }{1-\tan \alpha \times \tan (-\beta)}=\frac{\tan \alpha -\tan \beta }{1+\tan \alpha \times \tan \beta}}.
⯁ ⯁ ⯁
We can give the trigonometric functions of \latex{2\alpha} (double angle formulae) even more easily, as here we only have to use that \latex{\alpha+\alpha=2\alpha}.
THEOREM:
\latex{\sin 2\alpha =2\times \sin \alpha \times \cos \alpha};
\latex{\cos 2\alpha =\cos ^2\alpha -\sin ^2\alpha};
\latex{\tan 2\alpha =\frac{2\times \tan \alpha }{1-\tan ^2\alpha } },
\latex{\cos 2\alpha =\cos ^2\alpha -\sin ^2\alpha};
\latex{\tan 2\alpha =\frac{2\times \tan \alpha }{1-\tan ^2\alpha } },
where \latex{\cos 2\alpha \neq 0;\;\cos \alpha \neq 0;\;\tan ^2\alpha \neq 1}
\latex{\sin 2\alpha =2\times \sin \alpha \times \cos \alpha};
\latex{\cos 2\alpha =\cos ^2\alpha -\sin ^2\alpha};
\latex{\tan 2\alpha =\frac{2\times \tan \alpha }{1-\tan ^2\alpha } },
\latex{\cos 2\alpha =\cos ^2\alpha -\sin ^2\alpha};
\latex{\tan 2\alpha =\frac{2\times \tan \alpha }{1-\tan ^2\alpha } },
These relations make it possible to determine relations even relating to multiple angles.
Example 2
Let us express the value of sin \latex{3\alpha} in terms of the sine of the angle \latex{\alpha}.
Solution
Let us use that \latex{3\alpha=2\alpha+\alpha}, therefore:
\latex{\sin 3\alpha =\sin \left(2\alpha +\alpha \right)=\sin 2\alpha \times \cos \alpha +\cos 2\alpha \times \sin \alpha= }
\latex{=2\times \sin \alpha \times \cos ^2\alpha +\left(\cos ^2\alpha-\sin ^2\alpha \right)\times \sin \alpha=}
\latex{=3\times \sin \alpha \times \cos ^2\alpha -\sin ^3\alpha}.
\latex{=2\times \sin \alpha \times \cos ^2\alpha +\left(\cos ^2\alpha-\sin ^2\alpha \right)\times \sin \alpha=}
\latex{=3\times \sin \alpha \times \cos ^2\alpha -\sin ^3\alpha}.
Since \latex{\cos ^2\alpha =1-\sin ^2\alpha}:
\latex{\sin 3\alpha =3\times \sin \alpha \times \left(1-\sin ^2\alpha \right)-\sin ^3\alpha =3\times \sin \alpha -4\times \sin ^3\alpha}.
Example 3
The lengths of the sides of a triangle are consecutive whole numbers. The size of the largest angle is twice the size of the smallest one. Let us give the angles and the sides of the triangle.
Solution
Considering that there is a greater angle opposite a longer side let us denote the sides and the angles of the triangle as shown in the figure. (Figure 37)
According to the sine rule:
\latex{\frac{a+1}{a-1}= \frac{\sin 2\alpha }{\sin \alpha }=\frac{2\times \sin \alpha \times \cos \alpha }{\sin \alpha }=2\times \cos \alpha}.
\latex{\alpha}
\latex{2\alpha}
\latex{ a+1 }
\latex{ a-1 }
\latex{ B }
\latex{ C }
\latex{ A }
\latex{ a }
Figure 37
Let us write down the cosine rule for the side opposite the angle \latex{\alpha}:
\latex{\left(a-1\right)^2=a^2+\left(a+1\right)^2-2a\times \left(a+1\right)\times \cos \alpha}.
Let us substitute the value of \latex{2\cdot \cos\alpha} based on the sine rule and let us simplify the resulting equation:
\latex{\left(a-1\right)^2=a^2+\left(a+1\right)^2-a\times \left(a+1\right)\times \frac{a+1}{a-1}};
\latex{a^2-2a+1=a^2+a^2+2a+1-\frac{a\times \left(a+1\right)^2 }{a-1}};
\latex{a^2+4a=\frac{a\times \left(a+1\right)^2 }{a-1} }.
Since \latex{a\neq0}, we can divide both sides of the equation by \latex{ a }, and thus:
\latex{\left(a+4\right)\times \left(a-1\right)=\left(a+1\right)^2},
\latex{a^2+3a-4=a^2+2a+1},
a = 5.
\latex{a^2+3a-4=a^2+2a+1},
a = 5.
So the sides of the triangle are \latex{ 4 }, \latex{ 5 } and \latex{ 6 } units long.
For the angle \latex{\alpha} of the triangle:
\latex{\cos \alpha =\frac{a+1}{2\times \left(a-1\right) }=\frac{3}{4}}. From here: \latex{\alpha \approx 41.41°}
So the size of the angle opposite the longest side will be \latex{ 82.82º }, while – by using the relation relating to the interior angles of the triangle – the size of the third angle will be \latex{ 55.77º }.
If we determine the angles of a triangle with \latex{ 4 }, \latex{ 5 } and \latex{ 6 } units long sides with the help of one of the trigonometric theorems valid for triangles, then we get the same angles, which means that this triangle indeed meets the conditions of the example.
Example 4
Let us determine the trigonometric functions of the angle \latex{\alpha} if we know that \latex{\sin 2\alpha =\frac{1}{3}}.
Solution
According to the condition: \latex{2\times \sin \alpha \times \cos \alpha =\frac{1}{3}}.
Since the equality cannot be satisfied in the case of \latex{\cos\alpha=0}, we can divide by \latex{2\times \cos\alpha}:
\latex{\sin \alpha =\frac{1}{6\times \cos \alpha }.}
By substituting it into the known equality \latex{\sin ^2\alpha +\cos ^2\alpha =1}:
\latex{\frac{1}{36\times \cos ^2\alpha }+\cos ^2\alpha =1}.
We can multiply the equation by the denominator of the fraction:
\latex{36\times \cos^4\alpha -36\times \cos^2\alpha +1=0}.
By solving the resulting equation:
\latex{\cos ^2\alpha =\frac{3\pm 2\times \sqrt{2} }{6}}.
We have to accept both solutions, thus four distinct values can meet the given conditions:
\latex{\cos \alpha _1=\sqrt{\frac{3+2\times \sqrt{2} }{6} } }, \latex{\cos \alpha _2=-\sqrt{\frac{3+2\times \sqrt{2} }{6} } },
\latex{\cos \alpha _3=\sqrt{\frac{3-2\times \sqrt{2} }{6} } }, \latex{\cos \alpha _4=-\sqrt{\frac{3-2\times \sqrt{2} }{6} } }.
According to this four solutions are obtained for the value of \latex{\sin\alpha} too:
\latex{\sin \alpha _1=\frac{1}{6\times \sqrt{\frac{3+2\times \sqrt{2} }{6} } } =\frac{1}{\sqrt{18+12\times \sqrt{2} } } }, \latex{\sin\alpha_2=-\frac{1}{\sqrt{18+12\times \sqrt{2} } } },
\latex{\sin\alpha_3=\frac{1}{\sqrt{18-12\times \sqrt{2} } } }, \latex{\sin\alpha_4=-\frac{1}{\sqrt{18-12\times \sqrt{2} } } }.
Since \latex{\tan{\alpha}=\frac{\sin\alpha}{\cos\alpha}} thus by using the condition \latex{\sin\alpha=\frac{1}{6\times\cos\alpha}} of the example:
\latex{\tan{\alpha}=\frac{1}{6\times\cos^2\alpha}}.
From here the possible values of tana are:
\latex{\tan{\alpha_1}=\frac{1}{6\times\left(\pm\sqrt{\frac{3+2\times\sqrt2}{6}}\right)}=\frac{1}{3+2\times\sqrt2}=3-2\times\sqrt2},
and\latex{\tan{\alpha_2}=\frac{1}{6\times\left(\pm\sqrt{\frac{3-2\times\sqrt2}{6}}\right)}=\frac{1}{3-2\times\sqrt2}=3+2\times\sqrt2}.
Based on \latex{\cot\alpha=\frac{1}{\tan\alpha}} the possible values of \latex{\cot\alpha} are:
\latex{\cot\alpha_1=\frac{1}{3-2\sqrt2}=3+2\sqrt2}, \latex{\cot\alpha_2=\frac{1}{3+2\times\sqrt2}=3-2\times\sqrt2}.
We can observe that while even four possible values can be obtained for the sine and cosine of the angle a, only two values can be obtained for the value of the tangent and the cotangent. This is due to the quotient \latex{\frac{\sin\alpha}{\cos\alpha}}.
Example 5
At what angle shall we throw a stone if we want it to get the furthest away as possible on the level ground? (Let us ignore the air resistance.)
Solution
If a stone is thrown at an angle of a to the horizontal with an initial velocity of \latex{ v_0 }, then we know that its motion can be considered as the sum of two independent motions.
(Figure 38)
(Figure 38)
One of them is a horizontal steady linear motion, the velocity of which – \latex{v_0\times\cos\alpha} – stays constant, since no force is applied in this direction which would speed it up or slow it down.
\latex{\alpha}
\latex{v_0\times\cos\alpha}
\latex{v_0}
\latex{v_0\times\sin\alpha}
\latex{S_x}
Figure 38
However vertically it has a uniformly accelerated linear motion the initial velocity of which is \latex{v_0\cdot\sin\alpha}, and its acceleration will be \latex{ –g }. (For the acceleration we take into account that \latex{\vec{a}} and the initial velocity vector are in the opposite directions.)
If the motion happens for time \latex{ t }, then its vertical displacement will be \latex{ 0 }, since it will get back to the initial level:
\latex{0=v_0\times \sin \alpha \times t-\frac{g}{2}\times t^2}.
The flying time of the stone can be expressed from this equation in terms of the angle \latex{\alpha}:
\latex{t=\frac{2\times v_0\times \sin \alpha }{g} }.
Now we have to select the angle \latex{\alpha} so that the horizontal displacement of the stone will be the largest possible:
\latex{S_x=v_0\times \cos \alpha \times t=v_0\times \cos \alpha \times \frac{2\times v_0\times \sin \alpha }{g}= }
\latex{=\frac{v^2_0\times 2\times\sin \alpha \times \cos \alpha }{g}=\frac{v^2_0}{g}\times \sin 2\alpha }.
Based on the relation obtained it will come true if the value of sin \latex{2\alpha} is maximized, and it will happen – by considering the range of the sine function – if:
\latex{\sin2\alpha=1}.
It implies that \latex{2\alpha=90°}, i.e. \latex{\alpha=45°}.
So we can throw the stone the furthest possible away if we throw it away at an angle of \latex{ 45º }.
Exercises
{{exercise_number}}. Give the value of the tangent of the following angles without using a calculator or a function value table:
- \latex{15°};
- \latex{-75°};
- \latex{105°}.
{{exercise_number}}. Verify the addition theorems for the cotangent function:
- \latex{\cot \left(\alpha +\beta \right)=\frac{\cot \alpha \times \cot \beta -1}{\cot \alpha +\cot \beta} };
- \latex{\cot \left(\alpha -\beta \right)=\frac{\cot \alpha \times \cot \beta +1}{\cot \alpha -\cot \beta} }.
{{exercise_number}}. Prove that \latex{\cot 2\alpha =\frac{\cot ^2\alpha -1}{2\times \cot \alpha }}.
{{exercise_number}}. Determine the trigonometric functions of the angle \latex{\alpha} if we know that:
- \latex{\sin 2\alpha =\frac{1}{4} };
- \latex{\cos 2\alpha =\frac{1}{3}};
- \latex{\tan\alpha =\frac{1}{4}};
- \latex{\cot\alpha=2}.
{{exercise_number}}. Express the value of \latex{\cos3\alpha} in terms of the cosine of the angle \latex{\alpha}.
{{exercise_number}}. Verify the following identities:
- \latex{\frac{2}{\sin 2\alpha }=\tan \alpha +\cot \alpha };
- \latex{\frac{\sin 2\alpha }{\sin ^2\alpha }=2\times \cot \alpha };
- \latex{2\times \left[\cos \left(45°+\alpha \right) \right]^2=1-\sin 2\alpha .}
{{exercise_number}}. Two sides of a triangle are \latex{ 6\, cm } and \latex{ 7\, cm } long; the ratio of the angles opposite them is \latex{ 1 : 2 }. Give the angles and the unknown side of the triangle.
{{exercise_number}}. Two sides of a triangle are \latex{ 20\, cm } and \latex{ 24\, cm } long; the angle bisector of the angle included between them is \latex{ 22\, cm } long. What is the measure of the angle included between the two sides?