UNIT 1: ANALYTICAL METHODS FOR ENGINEERS

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1 UNIT 1: ANALYTICAL METHODS FOR ENGINEERS Unit code: A/601/1401 QCF Level: 4 Credit value: 15 OUTCOME - TRIGONOMETRIC METHODS TUTORIAL TRIGONOMETRIC IDENTITIES Sinusoidal functions: review of the trigonometric ratios; Cartesian and polar co-ordinate systems; properties of the circle; radian measure; usoidal functions Applications such as: angular velocity; angular acceleration; centripetal force; frequency; amplitude; phase; the production of complex waveforms ug usoidal graphical synthesis; AC waveforms and phase shift Trigonometric identities: relationship between trigonometric and hyperbolic identities; double angle and compound angle formulae and the conversion of products to sums and differences; use of trigonometric identities to solve trigonometric equations and simplify trigonometric expressions You should judge your progress by completing the self assessment exercises. Trigonometry has been covered in the NC Maths module and should have been studied prior to this module. This tutorial provides further studies and applications of that work. D.J.Dunn 1

2 1. REVIEW OF TRIGONOMETRIC RATIOS This is the work covered so far in previous tutorials. SINE COSINE (A) = a/c ec(a) = c/a (A) = b/c sec(θ) = c/b TANGENT tan(a) = a/b cot (A) = b/a This is also useful to know. tan (A) = (A) / (A) (A) = a /c and (A) = b /c and (A) + (A) = a /c + b /c a b (A) (A) c = a + b (Pythagoras) c a b (A) (A) 1 a b a b c SINE RULE A B C COSINE RULE (A) b c a bc c (B) a b ca a (C) b c ab. COMPOUND ANGLES SUMS and DIFFERENCES Prove that (A + B) = (A) (B) (A) (B) Refer to the diagram opposite. OP (A+B) = OR (A) + RP (A) substitute (A) = - (A) OP (A+B) = OR (A) - RP (A) OR = OP (B) and RP = OP (B) OP (A+B) = OP (B)(A) - OP (A) (B) (A+B) = (A) (B) - (A) (B) Prove that (A + B) = (A) (B) + (A) (B) Refer to the diagram opposite. OP (A+B) = OR (A) RP (A) -(A) = (A) OP (A+B) = OR (A) + RP (A) OR = OP (B) and RP = OP (B) OP (A+B) = OP (B) (A) + OP (B) (A) (A+B) = (A) (B) + (B) (A) By a similar manner to before it can be shown that: (A - B) = (A) (B) + (A) (B) and (A - B) = (A) (B) - (A) (B) D.J.Dunn

3 DOUBLE ANGLES (A+B) = (A) (B) - (A) (B) put B = A: (A) = (A)(A) (A) (A) (A) = (A) (A) and ce (A) + (A) = 1 (A) = 1 (A) = (A) 1 1 (A) (A) (A) 1 and (A) Similarly (A+B) = (A) (B) + (B) (A) (A) = (A) (A) or (A) (A) = ½ (A) HALF ANGLES In the identity (A) = (A) (A) if we substitute A = C/ for A we get (C) = (C/) (C/) (C) = (C/) {1 - (C/)} (C) = (C/) 1 And (C) = (C/) (C/) 3. PRODUCTS AND SUMS CHANGING PRODUCTS TO SUMS We have alrea shown that: (A + B) = (A) (B) + (A) (B) (A - B) = (A) (B) - (A) (B) If we add the two lines we get : (A + B) + (A - B) = (A) (B) Rearrange, (A) (B) = ½ { (A + B) + (A B)} If A = B then as before (A) (A) = ½ { (A) } CHANGING SUMS TO PRODUCTS Ug Change sides Let C = A + B and D = A - B (A) (B) = ½ { (A+B) + (A B)} ½ { (A+B) + (A B)} = (A) (B) ½ { (C) + (D)} = (A) (B) A = C B and A = D + B add them together and A = C + D and subtracting B = C D Since C and D are only symbols for angles it must be true that: ½ { (C) + (D)} = {½ (C+D)} {½ (C - D)} { (C) + (D)} = {½ (C+D)} {½ (C - D)} A + B = {½ (A + B)} {½ (A B)} Likewise we can show: A B = {½ (A B)} {½ (A + B) } A + B = {½ (A + B)} {½ (A B)} A B = {½ (A + B)} {½ (A B)} D.J.Dunn 3

4 3. CHANGING COS TO SIN AND SIN TO COS Here we see how to change expressions of the form a(θ) + b (θ) into a gle term of or. a b a(θ) + b (θ) = c θ θ c c For a right angle triangle as shown we know that: (A) = a/c (A) = b/c c We can substitute these terms and get: a b a (θ) + b (θ) = c{(a) (θ) + (A) (θ)} Now use the identity {(A) (θ) + (A) (θ)}= (A + θ) a (θ) + b (θ) = c {(A + θ)} If a = b = 1 then c = A = 45 o (θ) + (θ) = {(45 o + θ)} If we started with the expression a (θ) - b (θ) the result would be: a b a (θ) - b (θ) = c θ θ c c a (θ) - b (θ) = c{(a) (θ) - (A) (θ)} Now use the identity (A - θ) = (A) (θ) - (A) (θ) a (θ) - b (θ) = c {(A - θ)} If a = b = 1 then c = A = 45 o (θ) - (θ) = {(45 o - θ)} WORKED EXAMPLE No.1 Change 3(θ) 4(θ) into e form and then ine form. a (θ) - b (θ) = c{(a-θ) } Let a = 3 and b = 4 hence c = 5 = 5 3 (θ) - 4 (θ) = 5{(A - θ} (A) = b/c = 4/5 (A) = a/c = 3/5 A = -1 (4/5) = 36.9 o or -1 (3/5) = 36.9 o 3(θ) 4(θ) = 5 (36.87 θ) = 5 (θ ) To put the expression into ine form we only need to know that (α) = (90 o -α) Hence (θ ) = (90 - θ ) = (- θ 53.13) = (θ ) 3(θ) 4(θ) = 5(θ ) Check put θ = 45 o and 3(θ) 4(θ) = (θ ) = (θ ) = D.J.Dunn 4

5 WORKED EXAMPLE No. Show that 1 θ θ tanθ Substitute (θ) = (θ) (θ) Substitute (θ) = (θ) 1 Simplify and θ θ θ θ θ θ 1 1 θ θ θ θ θ θ 1 θ tanθ θ θ θ θ WORKED EXAMPLE No. 3 Given (θ) + 3(θ) = 1/ Solve θ Substitute (θ) = 1 - (θ) {1 - (θ)} + 3(θ) = 7/ - (θ)} + 3(θ) = 7/ - (θ)} + 3(θ) 3/ = 0 (θ) - 3(θ) - 3/ = (4)()(3/ ) ( θ) ()() 4 4 (θ) = or Since the ine value can not exceed 1 the only answer must be (θ) = θ = o WORKED EXAMPLE No. 4 A generator produces a voltage v = 00 (πft) and a current i = 5 (πft) into a purely resistive load. The electric power is P = vi. Express this as a gle term and show that the power varies from zero to 1000 Watts at double the frequency. P = {00 (πft)} {5 (πft)} = 1000 (πft) 1 (A) Use the double angle formula (A) 1 π f t P π f t The frequency is doubled and P varies 0 to 1000 (see the plot) D.J.Dunn 5

6 4. HYPERBOLIC FUNCTIONS This was covered in a previous tutorial and the comparison table is given again here. h(-a) = - h(a) h(-a) = h(a) Hyperbolic function (-A) = - (A) (-A) = (A) Trigonometric function h ( θ) h ( θ) 1 θ + θ = 1 h(a) = h(a) h(a) h(a) = h (A)- h (A) (A) = (A) (A) (A) = (A)- (A) (A) = (A/) 1 (A) = (A/) (A/) (A) (B) = ½ { (A+B) + (A B)} (A) (B) = ½ { (A+B) - (A B)} (A) (B) = ½ { (A+B) + (A B)} (A) (B) = -½ { (A+B) - (A B)} (θ) ± (θ) = {(45 o ± θ)} h(a±b) = h(a) h(b) ± h(a) h(b) h(a±b) = h(a) h(b) ± h(a) h(b) (A ± B) = (A) (B) ± (A) (B) (A ± B) = (A) (B) (A) (B) h(x) + h(x) = e x h(x) - h(x) = e -x y h(x) then h(x) y (x) then (x) If y h(x) then h(x) If y (x) then (x) h(x) h(x) C (x) - h(x) C h(x) h(x) C (x) (x) C D.J.Dunn 6

7 SELF ASSESSMENT EXERCISE No.1 1. Given (3x) = 0.5 solve the two smallest positive values of x. (Answer 10 o or 50 o ). The velocity at which a vehicle overturns on an inclined bend is given by v Rg Make θ the subject of the formula Determine the angle if the vehicle overturns at 13 m/s when R = 30 m and μ = 0. (Answer 18.6 o ) 3. Given (θ) + (θ) = 1.5 Solve the smallest positive value of θ (Answer 45 o ) 4. Prove the following by ug standard trigonometric identities. 1 μ tanθ μtanθ i. 1 θ ( θ) tanθ θ ii. tan 1 ( θ) 1 θ 5. Show that tana B tana tanb 1 tana tanb 6. Change 5(α) + (α) into ine form (Answer (α -1.8 o ) 7. Change 3 ( θ) ( θ) into e form. (Answer (θ - 30 o ) 8. Two alternating voltages are expressed as v 1 = 3 (4t) and v = 4 (4t) When the voltage is summed the result is v = V (4t + ). Determine the value of V and (5 and o ) 9. Given that (θ) + (θ) = x find a formula with θ as the subject. Given x =.065 solve the smallest positive value of θ. (Answer 4.4 o ) 10. In a complex stress situation, the stress on a plane at angle θ to the x plane is given by the formula: σx σy σx σy σθ ( θ) τ(θ) Where The stress on the x plane is σ x = 00 MPa The stress on the y plane is σ y = 100 MPa The shear stress is τ = 50 MPa Calculate the angle of the plane where the stress is 18.3 MPa. (Answer 15 o ) D.J.Dunn 7