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MATHEMATICS FOR ENGINEERING

TRIGONOMETRY

TUTORIAL 4 – TRIGONOMETRIC IDENTITIES This is the one of a series of basic tutorials in mathematics aimed at beginners or anyone wanting to refresh themselves on fundamentals. The tutorial revises and extends the work on trigonometrical formula contains the following.

• Revision of trigonometrical identities • Double angle formulae • Compound angle formulae • Products, sums and differences

• The relationship between trigonometrical and hyperbolic identities.

1. REVIEW OF TRIGONOMETRIC RATIOS This is the work covered so far in previous tutorials. SINE sin(A) = b/c cosec(A) = c/b COSINE cos(A) = a/c sec(θ) = c/a TANGENT tan(A) = b/a cot (A) = a/b This is also useful to know. tan (A) = sin(A) / cos(A) sin2(A) = b2/c2 and cos2(A) = a2/c2 and sin2(A) + cos2(A) = b2/c2 + a2/c2

2

2222

cab(A)cos(A)sin +

=+ c2 = a2 + b2 (Pythagoras)

1baab(A)cos(A)sin 22

2222 =

++

=+

SINE RULE sinC

csinB

bsinA

a==

COSINE RULE

2bcacbcos(A)

222 −+=

2cabaccos(B)

222 −+=

2abcbacos(C)

222 −+=

2. COMPOUND ANGLES SUMS and DIFFERENCES Prove that cos(A + B) = cos(A) cos(B) – sin(A) sin(B) Refer to the diagram opposite. OP cos(A+B) = OR cos(A) + RP cos(A) substitute cos(A) = - sin(A) OP cos(A+B) = OR cos(A) - RP sin(A) OR = OP cos(B) and RP = OP sin(B) OP cos(A+B) = OP cos(B)cos(A) - OP sin(A) sin(B) cos(A+B) = cos(A) cos(B) - sin(A) sin(B) Prove that sin(A + B) = sin(A) cos(B) + cos(A) sin(B) Refer to the diagram opposite. OP sin(A+B) = OR sin(A) – RP sin(A) -sin(A) = cos(A) OP sin(A+B) = OR sin(A) + RP sin(A) OR = OP cos(B) and RP = OP sin(B) OP sin(A+B) = OP cos(B) sin(A) + OP sin(B) cos(A) sin(A+B) = sin(A) cos(B) + sin(B) cos(A) By a similar manner to before it can be shown that: cos(A - B) = cos(A) cos(B) + sin(A) sin(B) and sin(A - B) = sin(A) cos(B) - cos(A) sin(B)

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DOUBLE ANGLES cos(A+B) = cos(A) cos(B) - sin(A) sin(B) put B = A: cos(2A) = cos(A)cos(A) – sin(A) sin(A) cos(2A) = cos2(A) – sin2(A) and since sin2(A) + cos2(A) = 1 cos(2A) = 1 – 2sin2(A) = 2cos2(A) – 1

2

cos(2A)1(A)sin2 −=

and 2

1-cos(2A)(A)cos2 =

Similarly sin(A+B) = sin(A) cos(B) + sin(B) cos(A) sin(2A) = 2sin(A) cos(A) or sin(A) cos(A) = ½ sin(2A) HALF ANGLES In the identity cos(2A) = cos2(A) – sin2(A) if we substitute A = C/2 for A we get cos(C) = cos2(C/2) – sin2(C/2) cos(C) = cos2(C/2) – {1 - cos2(C/2)} cos(C) = 2cos2(C/2) – 1 And sin(C) = 2sin(C/2) cos(C/2) 3. PRODUCTS AND SUMS CHANGING PRODUCTS TO SUMS We have already shown that: sin(A + B) = sin(A) cos(B) + cos(A) sin(B) sin(A - B) = sin(A) cos(B) - cos(A) sin(B) If we add the two lines we get : sin(A + B) + sin(A - B) = 2sin(A) cos(B) Rearrange, sin (A) cos (B) = ½ {sin (A + B) + sin (A − B)} If A = B then as before sin (A) cos (A) = ½ {sin (2A) } CHANGING SUMS TO PRODUCTS Using sin (A) cos (B) = ½ {sin (A+B) + sin (A−B)} Change sides ½ {sin (A+B) + sin (A−B)} = sin (A)cos (B) Let C = A + B and D = A - B ½ {sin (C) + sin (D)} = sin (A) cos (B) A = C – B and A = D + B add them together and 2A = C + D and subtracting 2B = C – D ½ {sin (C) + sin (D)} = sin {½ (C+D)} cos {½ (C - D)} {sin (C) + sin (D)} = 2 sin {½ (C+D)} cos {½ (C - D)} Since C and D are only symbols for angles it must be true that: sin A + sin B = 2 sin {½ (A + B)} cos {½ (A − B)} Likewise we can show: sin A − sin B = 2 sin {½ (A − B)} cos {½ (A + B) } cos A + cos B = 2 cos {½ (A + B)} cos {½ (A − B)} cos A − cos B = −2 sin {½ (A + B)} sin {½ (A − B)}

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3. CHANGING COS TO SIN AND SIN TO COS For a right angle triangle as shown we know that: sin(A) = b/c cos(A) = a/c 22 bac += Consider the expression a cos (θ) + b sin(θ) (noting θ is not A)

This is the same as the expression ( ) ( )θsincbcθcos

cac

⎭⎬⎫

⎩⎨⎧+

⎭⎬⎫

⎩⎨⎧

a cos (θ) + b sin(θ) = ( ) ( )⎥⎦

⎤⎢⎣

⎡⎭⎬⎫

⎩⎨⎧+

⎭⎬⎫

⎩⎨⎧ θsin

cbθcos

cac

a cos (θ) + b sin(θ) = c{cos(A) cos(θ) + sin(A) sin(θ)} Now use the identity cos(A - B) = cos(A) cos(B) + sin(A) sin(B) Only in this case it is cos(A - θ) = cos(A) cos(θ) + sin(A) sin(θ) a cos (θ) + b sin(θ) = c{cos(A – θ)} If a = b = 1 then c = √2 A = 45o cos (θ) + sin(θ) = √2 {cos(45o – θ)} If we started with the expression b cos (θ) - a sin(θ)

b cos (θ) - a sin(θ) = ( ) ( )⎥⎦

⎤⎢⎣

⎡⎭⎬⎫

⎩⎨⎧−

⎭⎬⎫

⎩⎨⎧ θsin

caθcos

cbc

b cos (θ) - a sin(θ) = c{sin(A) cos(θ) - cos(A) sin(θ)} Now use the identity sin(A - θ) = sin(A) cos(θ) - cos(A) sin(θ) b cos (θ) - a sin(θ) = c{sin(A - θ)} If a = b = 1 then c = √2 A = 45o cos (θ) - sin(θ) = √2{sin(45o - θ)} WORKED EXAMPLE No.1 Change 3cos(θ) – 4sin(θ) into cosine form and then sine form. SOLUTION b cos (θ) - a sin(θ) = c{sin(A-θ) } Let b = 3 and a = 4 hence c = √25 = 5 3 cos (θ) - 4 sin(θ) = 5{sin(A - θ} cos(A) = a/c = 3/5 sin(A) = b/c = 4/5 α = cos-1(3/5) = 53.13o or sin-1(4/5) = 53.13o

)53.13θcos(5θ) 4sin(θ) 3cos( o+=−

Check put θ = 45o 707.0θ) 4sin(θ) 3cos( −=− and -0.707)53.13θcos(5 o =+ To put the expression into sine form we only need to know that cos(A) = sin(90o- A) Hence cos(θ + 53.13o) = sin(90 - θ - 53.13) = sin(36.87- θ) 3cos(θ) – 4sin(θ) = sin(36.87- θ) The sin is repeated at ±180o so it might be tidier to write 3cos(θ) – 4sin(θ) = sin(θ + 143.13o) Check put θ = 45o -0.707)13.431θsin(5 o =+

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WORKED EXAMPLE No.2

Show that ( )( ) ( )θtan2θcos1

2θsin=

+

SOLUTION

Substitute sin(2θ) = 2sin(θ) cos(θ) ( )( )

( ) ( )( )2θcos1

θcosθ2sin2θcos1

2θsin+

=+

Substitute cos(2θ) = 2cos2(θ) – 1 ( )( )

( ) ( )( )

( ) ( )( )θcos2

θcosθ2sin1θcos21

θcosθ2sin2θcos1

2θsin22 =

−+=

+

Simplify and ( )( )

( )( ) ( )θtanθcosθsin

2θcos12θsin

==+

WORKED EXAMPLE No. 3 Given 2sin2(θ) + 3cos(θ) = 1/2 Solve θ SOLUTION Substitute sin2(θ) = 1 - cos2(θ) 2{1 - cos2(θ)} + 3cos(θ) = 7/2 2 - 2cos2(θ)} + 3cos(θ) = 7/2 - 2cos2(θ)} + 3cos(θ) – 3/2 = 0 2cos2(θ) - 3cos(θ) - 3/2 = 0

4

213 4

1293)2)(2(

)2/3)(2)(4(33θ) cos(

2 ±=

+±=

+±=

cos(θ) = 1.895 or -0.396 Since the cosine value can not exceed 1 the only answer must be cos(θ) = -0.396 θ = 113.3o WORKED EXAMPLE No. 4 A generator produces a voltage v = 200 sin(2πft) and a current i = 5 sin(2πft) into a purely

resistive load. The electric power is P = vi. Express this as a single term and show that the power varies from zero to 1000 Watts at double the frequency.

SOLUTION

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P = {200 sin(2πft)} {5 sin(2πft)} = 1000 sin2(2πft)

Use the double angle formula 2

cos(2A)1(A)sin2 −=

( ){ } { } tf π4cos5000052

tf π22cos11000P −=⎟⎠⎞

⎜⎝⎛ −

=

The frequency is doubled and P varies 0 to 1000 (see the plot)

4. HYPERBOLIC FUNCTIONS This was covered in a previous tutorial and the comparison table is given again here.

Hyperbolic function Trigonometric function

sinh(-A) = - sinh(A) sin(-A) = - sin(A)

cosh(-A) = cosh(A) cos(-A) = cos(A)

1θ) (sinhθ) (cosh 22 =+ sin2θ + cos2θ = 1

sinh(2A) = 2sinh(A) cosh(A) sin(2A) = 2sin(A) cos(A)

cosh(2A) = cosh2(A)- sinh2(A) cos(2A) = cos2(A)- sin2(A)

cos(A) = 2cos2(A/2) – 1

sin(A) = 2sin(A/2) cos(A/2)

sin (A) cos (B) = ½ {sin (A+B) + sin (A−B)}

cos (A) sin (B) = ½ {sin (A+B) - sin (A−B)}

cos (A) cos (B) = ½ {cos (A+B) + cos (A−B)}

sin (A) sin (B) = -½ {cos (A+B) - cos (A−B)}

cos (θ) - sin(θ) = √2{sin(45o - θ)}

cos (θ) + sin(θ) = √2 {cos(45o – θ)}

sinh(A±B) = sinh(A) cosh(B) ± cosh(A) sinh(B) sin(A ± B) = sin(A) cos(B) ± cos(A) sin(B)

cosh(A±B) = cosh(A) cosh(B) ± sinh(A) sinh(B) cos(A ± B) = cos(A) cos(B) sin(A) sin(B) m

cosh(x) + sinh(x) = ex

cosh(x) - sinh(x) = e-x

cosh(x)dxdy then sinh(x)y == cos(x)

dxdy then sin(x)y ==

sinh(x)dxdy then cosh(x) y If == sin(x)

dxdy then cos(x) y If −==

Ccosh(x) sinh(x)dx +=∫ Ccosh(x)- sin(x)dx +=∫

∫ += Csinh(x)cosh(x)dx ∫ += Csin(x)cos(x)dx

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SELF ASSESSMENT EXERCISE No.1 1. Given sin(3x) = 0.5 solve the two smallest positive values of x. (Answer 10o or 50o)

2. The velocity at which a vehicle overturns on an inclined bend is given by { }{ }µtanθ1

tanθµRgv2

−+

=

Make θ the subject of the formula Determine the angle if the vehicle overturns at 13 m/s when R = 30 m and µ = 0.2 (Answer 18.6o) 3. Given cos2(θ) + 2sin2(θ) = 1.5 Solve the smallest positive value of θ (Answer 45o) 4. Prove the following by using standard trigonometric identities.

i. ( ) ( )θtanθ) sin(22θcos1

=− ii. ( ) ( )θtan

θ) cos(2 12θcos1 2=

+−

5. Show that ( )tanA tanB1

tanBtanABAtan−

+=+

6. Change 5cos(α) + 2 sin(α) into cosine form (Answer 5.385 cos(α -21.8o) 7. Change θ) cos(θ) sin( 3 − into cosine form. (Answer 2 sin(α -30o) 8. Two alternating voltages are expressed as v1 = 3 sin(4t) and v2 = 4 cos(4t) When the voltage is summed the result is v = V sin(4t + φ). Determine the value of V and φ (5 and 52o) 9. Given that 2cos2(θ) + sin(θ) = x find a formula with θ as the subject. Given x = 2.065 solve the smallest positive value of θ. (Answer 25o) 10. In a complex stress situation, the stress on a plane at angle θ to the x plane is given by the

formula:

τsin(2θ)θ) cos(22σσ

2σσ

σ yxyxθ +

−+

+=

Where The stress on the x plane is σx = 200 MPa The stress on the y plane is σy = 100 MPa The shear stress is τ = 50 MPa Calculate the angle of the plane where the stress is 218.3 MPa. (Answer 15o)

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