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Trigonometry

Toolkit

Name Formula
Sine Law

\(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R\), where \(R\) is the radius of the circumcircle.

Triangle ABC with its vertices on a circle of radius R. AB has length c, BC has length a, AC has length b.
Cosine Law \(a^2=b^2+c^2-2bc\cos A\),
\(b^2=a^2+c^2-2ac\cos B\),
\(c^2=b^2+a^2-2ab\cos C\)
Area relations The area of triangle \(ABC\) is \(\dfrac{1}{2}ab\sin C=\dfrac{1}{2}bc\sin A=\dfrac{1}{2}ac \sin B\).
General Identities \(\cot\theta=\dfrac{1}{\tan\theta}\), \(\sec\theta=\dfrac{1}{\cos\theta}\), \(\csc\theta=\dfrac{1}{\sin\theta}\),
\(\tan\theta = \dfrac{\sin \theta}{\cos \theta}\), \(\cot \theta = \dfrac{ \cos \theta} { \sin \theta}\),
\(\sin(-\theta) = - \sin \theta\), \(\cos(-\theta) = \cos \theta\), \(\tan(-\theta) = - \tan \theta\)
Pythagorean Identities \(\sin^2\theta+\cos^2\theta=1\), \(\tan^2\theta+1=\sec^2\theta\), \(\cot^2\theta+1=\csc^2\theta\)
Sum Formulas \(\sin(A + B) = \sin A \cos B + \cos A\sin B\),
\(\cos(A+B) = \cos A \cos B - \sin A \sin B\)
Double Angle Formulas \(\sin(2\theta) = 2 \sin \theta \cos \theta\),
\(\cos(2\theta) = \cos^2 \theta - \sin^2 \theta = 2\cos^2\theta -1 =1 -2 \sin^2\theta\)

Sample Problems

  1. Determine the values of \(x\) such that \(2\sin^3x-5\sin^2x+2\sin x=0\) given that \(0\le x\le2\pi\).

    Solution

    We factor the given equation to obtain \[\begin{align*} \sin x(2\sin^2x-5\sin x+2)&=0\\ \sin x(2\sin x-1)(\sin x-2)&=0\end{align*}\] So \(\sin x=0\), \(\dfrac{1}{2}\) or \(2\). But \(|\sin x|\le1\). So \(\sin x \neq 2\). Therefore, in the interval \(0\le x\le2\pi\), we have \(x=0\), \(\pi\), \(2\pi\), \(\dfrac{\pi}{6}\) or \(\dfrac{5\pi}{6}\).

  2. An airplane leaves an aircraft carrier and flies due south at \(400\) km/hr. The carrier proceeds at a heading of \(60\degree\) west of north at \(32\) km/hr. If the plane has \(5\) hours of fuel, what is the maximum distance south the plane can travel so that the fuel remaining will allow a safe return to the carrier at \(400\) km/hr?

    Solution

    The first step in solving this problem is to draw a diagram (as shown). If we let \(x\) be the number of hours that the plane flies south, then the distance that the plane flies south is \(400x\). The plane then flies a distance \(400(5-x)\) in the remaining time, while the total distance the carrier travels is \(5(32)=160\).

    A triangle plotted in the plane. The first vertex is at the origin. The second vertex is 400 times x units below the origin. The third vertex is in the second quadrant, 160 units from the first vertex and 400 times the quantity 5 minus x units from the second vertex.

    Using these distances, the cosine law gives us \[(400(5-x))^2=160^2+(400x)^2-2\cdot160\cdot400x\cdot\cos120\degree.\] Simplifying we obtain \[4000000-1600000x+160000x^2 = 25600+160000x^2+64000x\] which we can solve to get \(x=\dfrac{621}{260}\). Thus, the maximum distance the plane should travel south is \(400\left(\dfrac{621}{260}\right) = \dfrac{12420}{13}\) km, which is approximately \(955.4\) km.

  3. In triangle \(ABC\), the point \(D\) is on \(BC\) such that \(AD\) bisects \(\angle A\).

    Show that \(\dfrac{AB}{BD}=\dfrac{AC}{CD}\).

    Solution

    We call \(\angle ADC=\theta\) and \(\angle BAC = \alpha\). We use the sine law in triangles \(ADC\) and \(ADB\) to obtain \(\dfrac{\sin\dfrac{\alpha}{2}}{\sin\theta}=\dfrac{CD}{AC}\) and \(\dfrac{\sin\dfrac{\alpha}{2}}{\sin(180\degree-\theta)}=\dfrac{BD}{AB}\). But \(\sin\theta=\sin(180\degree-\theta)\) and so \(\dfrac{AB}{BD}=\dfrac{AC}{CD}\). This result is known as the angle bisector theorem.

  4. For the given triangle \(ABC\), \(\angle C=\angle A+60\degree\). If \(BC=1\), \(AC=r\) and \(AB=r^2\), where \(r>1\), prove that \(r<\sqrt{2}.\)

    Solution

    We represent the angles of the triangle as: \(\angle A=\alpha\), \(\angle C=\alpha+60\degree\), and \(\angle B=120\degree-2\alpha\). So the sine law states \[\begin{align*} \dfrac{r^2}{1}&=\dfrac{\sin(\alpha+60\degree)}{\sin\alpha}\\ &=\dfrac{\sin\alpha\cos60\degree+\cos\alpha \sin60\degree}{\sin\alpha}\\ &=\dfrac{1}{2} +\dfrac{\sqrt{3}}{2}\cot\alpha\end{align*}\] Since all three angles in the triangle are positive, we can see that \(0\degree<\alpha<60\degree\). In this range, the tangent function is increasing, and its reciprocal, the cotangent function, is decreasing.
    The cosine law gives \[r^2=1+r^4-2r^2\cos(120\degree-2\alpha).\] But \(r >1\) and so \[(r^2-1)^2 > 0 \textrm{ or } r^4+1 > 2r^2.\] Substituting the second inequality into the equation gives \(r^2 > 2r^2-2r^2\cos(120\degree-2\alpha)\) which implies \(\cos(120\degree-2\alpha) > \dfrac{1}{2}\). Thus, \(\alpha > 30\degree\) and

    \[r^2=\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\cot\alpha < \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{3}}{1} = 2\] Thus, \(r^2 < 2\) and so \(r < \sqrt{2}.\)

Problem Set

    1. If \(2\sin(2\theta)+1=0\), find the smallest positive value of \(\theta\) (in degrees).

    2. For \(-\pi\le\theta\le\pi\), find all solutions to the equation \(2(\sin^2\theta-\cos^2\theta)=8\sin\theta-5\).

  2. In \(\triangle ABC\), \(M\) is a point on \(BC\) such that \(BM=5\) and \(MC=6\). If \(AM=3\) and \(AB=7\), determine the exact value of \(AC\).

  3. In determining the height, \(MN\), of a tower on an island, two points \(A\) and \(B\), \(100\) m apart, are chosen on the same horizontal plane as \(N\) (the base of the tower). If \(\angle NAB=108\degree\), \(\angle ABN=47\degree\) and \(\angle MBN=32\degree\), determine the height of the tower to the nearest metre.

  4. A rectangle \(PQRS\) has side \(PQ\) on the \(x\)-axis and touches the graph of \(y=k\cos x\) at the points \(S\) and \(R\) as shown.

    P is on the negative x-axis and S is above P and on the graph. Q is on the positive x-axis and R above Q and on the graph. The curve crosses the x-axis at Pi over 2 and negative Pi over 2.

    If the length of \(PQ\) is \(\dfrac{\pi}{3}\) and the area of the rectangle is \(\dfrac{5\pi}{3}\), what is the value of \(k\)?

  5. The graph of the equation \(y=a\sin kx\) is shown in the diagram, and the point \(\bigg(\dfrac{3\pi}{4}\),\({-2}\bigg)\) is the minimum point indicated. The line \(y=1\) intersects the graph at point \(D\). What are the coordinates of \(D\)?

    The point (3 Pi over 4, negative 2) is the first local minimum of the function appearing to the right of the y-axis. D is located on the graph between the origin and the first local maximum of the function in the first quadrant.

  6. A square with an area of \(9\text{ cm}^2\) is surrounded by four congruent triangles, forming a larger square with an area of \(89\text{ cm}^2\). If each of the triangles has an angle \(\theta\) as shown, find the value of \(\tan\theta\).

    One triangle has its longest side as a horizontal base and bottom-right vertex with angle theta. Its base forms the bottom side of the larger square. Three copies of this triangle are rotated and arranged so that their longest sides form the other three sides of the larger square. The triangles do not overlap and fit together in such a way that they cover the entire area of the larger square except for a smaller tilted square in the middle.

  7. A rectangular box has a square base of length \(1\) cm, and height \(\sqrt{3}\) cm as shown in the diagram.

    Square ABCD forms the base of the box. Adjacent edges AB and BC have length 1. Square EHGF forms the top of the box with E above A, H above B, G above C, and F above D. Edge AE has length equal to the square root of 3.

    What is the cosine of angle \(FAC\)?

  8. In the grid, each small equilateral triangle has side length \(1\).

    Identical triangles are arranged to form a tiling. Each row in the grid alternates between triangles pointing up and triangles pointing down, placed so adjacent triangles share a side. Odd numbered rows start pointing down and even numbered rows start pointing up. In the first row, the bottom-right vertex of the sixth triangle is labelled W. In the fourth row, the bottom right vertex of the first triangle is labelled A. In the fifth row, the bottom-right vertex of the eighth triangle is labelled T.

    If the vertices of \(\triangle WAT\) are themselves vertices of small equilateral triangles, what is the area of \(\triangle WAT\)?

  9. In \(\triangle ABC\), \(AB=8\), and \(\angle CAB=60\degree\). Sides \(BC\) and \(AC\) have integer lengths \(a\) and \(b\), respectively. Find all possible values of \(a\) and \(b\).