Figure 5.1. sin θ = AB. cos θ = OB. tan θ = AB OB = sin θ. sec θ = 1. cotan θ = 1

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1 5 Trigonometric functions Trigonometry is the mathematics of triangles. A right-angle triangle is one in which one angle is 90, as shown in Figure 5.1. The thir angle in the triangle is φ = (90 θ). Figure 5.1 Six ratios can be constructe involving the three sies of a right-angle triangle an these epen only on the angle θ. The ratios are functions of the variable θ an are calle trigonometric functions. We have alreay assume familiarity with the basic trigonometric functions, sine, cosine an tangent but we list below all six for completeness. sin θ = AB OA, (5.1) cos θ = OB OA, (5.) tan θ = AB OB = sin θ cos θ, (5.3) cosec θ = 1 sinθ, (5.4) sec θ = 1 cosθ, (5.5) cotan θ = 1 tanθ. (5.6) The length of the hypotenuse is always positive an the signs of the lengths of the sies encompassing the right angle epen in the normal way on which sie of the point O the points A an B lie. With this convention the sines of angles between 90 an 180 are positive. For an angle π θ greater than 90, Figure 5., in which the length AB equals the length A B an the length OB equals the 1

2 length OB, shows that sin(π θ) =A B /OA =AB/OA = sin θ. The sines of angles between 180 an 360 are negative, an the signs of cosines are negative between 90 an 70, becoming positive between 70 an 360. As the angle θ is increase to 90, sin θ approaches unity. Figure 5. Example 5.1 Determine sin 10, cos 10 an tan 330. Using figures similar to 5.1 with sies,1 an 3, sin 10 = sin 30 = 1/, cos 10 = 1/ an tan 330 = 1/ Trigonometric relationships Many relationships can be erive between the trigonometric functions using Eucliian geometry. For example, Pythagoras theorem tells us that the square of the hypotenuse in a right-angle triangle equals the sum of the squares of the other two sies, an this immeiately leas to the relation for any angle θ. sin θ + cos θ = 1, (5.7) The following equations relating the angles θ an φ may also be prove. sin(θ ± φ) = sin θ cos φ ± cos θ sin φ, (5.8) cos(θ ± φ) = cos θ cos φ sin θ sin φ, (5.9) where θ an φ are any two angles, an means for cos(θ + φ) use the minus sign on the right-han sie, an for cos(θ φ) use the plus sign.

3 Many useful relations between trigonometric functions can be obtaine using equations 5.8 an 5.9 an simple algebraic manipulation. Problem 5.1 Use equations 5.8 an 5.9 to etermine sin 10, cos 10 an tan 330. Example 5. Show that sin θ = tan(θ/) 1 + tan (θ/). sin θ = sin(θ/ + θ/) = sin(θ/) cos(θ/) = sin(θ/) cos(θ/) sin (θ/) + cos (θ/), from equations 5.8 an 5.7. Diviing top an bottom by cos (θ/) gives the answer. Problem 5. Show that tan θ = tan(θ/) 1 tan (θ/). Problem 5.3 Using suitable constructions it is straightforwar to show that if a triangle has sies a, b an c an the angles opposite the sies are α, β an γ, a sin α = b sin β = c sin γ. Use this as the starting point an erive equation Differentials an Integrals Use can be mae of the above equations (5.8) an (5.9) to etermine the ifferentials an hence the integrals of the trigonometric functions. θ 1 (sin θ) = (sin(θ + δθ) sin θ), δθ in the limit as δθ tens to the vanishingly small θ. Hence, from equation (5.8), an θ (sin θ) = 1 θ (sin θ cos θ + cos θ sin θ sin θ), (sin θ) = cos θ. (5.10) θ To obtain the last equation we have use the fact that as δθ becomes the infinitesimally small θ, sin θ becomes θ an cos θ becomes unity, the cosine of a vanishingly small number. 3

4 In a similar way, using equation (5.8), it can be shown that θ (cos θ) = sin θ, (5.11) an θ (tanθ) = sec θ, (5.1) where the last relation can be erive using equation (5.3). Example 5.3 Show that θ (sin3 θ cos θ) = sin θ(4 cos θ 1). θ (sin3 θ cos θ) = 3 sin θ cos θ sin 4 θ = sin θ(3 cos θ sin θ) The solution use equations.10, 5.10 an = sin θ(4 cos θ 1). Problem 5.4 Show that θ ( ) sin θ cos θ = cos θ sin θ sin 3 θ 1 + sin θ (1 + sin θ). The integrals of the trigonometric functions can be obtaine as the reverse of the ifferentials, giving sin θ θ = cos θ, (5.13) cos θ θ = sin θ, (5.14) an tan θ θ = ln( cosθ ) = ln( secθ ). (5.15) The last relation is not so obvious as the first two but can reaily be verifie by ifferentiation. Several ifferentials an integrals of trigonometric expressions are given in the formulae section. One technique for performing integrals is the metho of substitution of variables first mentione in Section

5 Example 5.4 Integrate sin θ cos θθ. sin θ = cos θ, θ an (sin θ) = cos θθ. Hence ignoring the arbitrary constant. sin θ cos θθ = sin θ (sin θ) = 1 3 sin3 θ, Problem 5.5 Show that sin θ cos 3 θθ = sin3 θ cos θ sin3 θ. Trigonometric functions as exponentials The exponential function e jθ where θ is a real number is complex. The square of its moulus, an thus its moulus, is unity. The complex number e jθ e jθ = 1. z = cos θ + j sin θ, also has unit moulus, an it can be shown that e jθ = cos θ + j sin θ. (5.16) This connection between the exponential function with imaginary argument an sines an cosines is extremely useful. Since sin( θ) = sin(0 θ) = sin θ, from equation (5.8), an cos( θ) = cos(0 θ) = cos θ from equation 5.9, Using the last two equations gives e jθ = cos θ j sin θ. (5.17) sin θ = 1 j (ejθ e jθ ), (5.18) an cos θ = 1 (ejθ + e jθ ) (5.19) 5

6 Example 5.5 Show that equations 5.16 an 5.17 satisfy equation 5.8, thus establishing the valiity of that equation. sin(θ + φ) = 1 j (ej(θ+φ) e j(θ+φ) ). sin θ cos φ = 1 4j (ejθ e jθ )(e jφ + e jφ ) = 1 4j (ej(θ+φ) + e j(θ φ) e j(θ φ) e j(θ+φ) ). cos θ sin φ = 1 4j (ejθ + e jθ )(e jφ e jφ ) = 1 4j (ej(θ+φ) e j(θ φ) + e j(θ φ) e j(θ+φ) ). Hence sin θ cos φ + cos θ sin φ = 1 j (ej(θ+φ) e j(θ+φ) ) = sin(θ + φ). Problem 5.6 Show that equations 5.16 an 5.17 satisfy equation 5.9. Problem 5.7 Show that ( ) θ + φ sin θ + sin φ = sin cos an that ( ) θ + φ cos θ + cos φ = cos cos ( θ φ ( θ φ ), ). Polar form of complex numbers Using equation (5.16) we may now write the complex number z = a + jb in its polar form. If a complex number z = a + jb has a moulus r, z = (a + jb) = re jθ (5.0), where r is given by equation 4.3 an the argument θ is given by equation 4.4. It is often useful to express complex numbers in polar form, when multiplication becomes aition of the arguments of exponentials. Problem 5.8 Express z = ( 3j)(1 + j)/(4 + 3j) in polar form. 6

Math 123 Discussion Session Week 4 Notes April 25, 2017

Math 123 Discussion Session Week 4 Notes April 25, 2017 Math 23 Discussion Session Week 4 Notes April 25, 207 Some trigonometry Today we want to approach trigonometry in the same way we ve approached geometry so far this quarter: we re relatively familiar with