Download mathematics for engineering trigonometry tutorial 4

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.
© D.J.Dunn www.freestudy.co.uk
1
1.
REVIEW OF TRIGONOMETRIC RATIOS
This is the work covered so far in previous tutorials.
SINE
sin(A) = b/c
COSINE
cos(A) = a/c sec(θ) = c/a
TANGENT
tan(A) = b/a
This is also useful to know.
cosec(A) = c/b
cot (A) = a/b
tan (A) = sin(A) / cos(A)
sin2(A) = b2/c2 and cos2(A) = a2/c2 and sin2(A) + cos2(A) = b2/c2 + a2/c2
b2 + a 2 2
c = a2 + b2 (Pythagoras)
sin 2 (A) + cos 2 (A) =
2
c
2
b + a2
sin 2 (A) + cos 2 (A) = 2
=1
a + b2
a
b
c
SINE RULE
=
=
sinA sinB sinC
COSINE RULE
b2 + c2 − a 2
c2 + a 2 − b2
cos(A) =
cos(B) =
2bc
2ca
a 2 + b2 − c2
cos(C) =
2ab
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)
©D.J.Dunn www.freestudy.co.uk 2
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
1 − cos(2A)
sin 2 (A) =
2
cos(2A)
-1
cos 2 (A) =
and
2
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:
If we add the two lines we get :
Rearrange,
If A = B then as before
sin(A + B) = sin(A) cos(B) + cos(A) sin(B)
sin(A - B) = sin(A) cos(B) - cos(A) sin(B)
sin(A + B) + sin(A - B) = 2sin(A) cos(B)
sin (A) cos (B) = ½ {sin (A + B) + sin (A − B)}
sin (A) cos (A) = ½ {sin (2A) }
CHANGING SUMS TO PRODUCTS
Using
Change sides
Let C = A + B and D = A - B
sin (A) cos (B) = ½ {sin (A+B) + sin (A−B)}
½ {sin (A+B) + sin (A−B)} = sin (A)cos (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:
Likewise we can show:
©D.J.Dunn www.freestudy.co.uk 3
sin A + sin B = 2 sin {½ (A + B)} cos {½ (A − B)}
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)}
3. CHANGING COS TO SIN AND SIN TO COS
For a right angle triangle as shown we know that:
sin(A) = b/c
c = a 2 + b2
cos(A) = a/c
Consider the expression a cos (θ) + b sin(θ) (noting θ is not A)
⎧b⎫
⎧a ⎫
This is the same as the expression c⎨ ⎬cos(θ ) + c⎨ ⎬sin (θ )
⎩c⎭
⎩c⎭
⎤
⎡⎧ a ⎫
⎧b⎫
a cos (θ) + b sin(θ) = c ⎢⎨ ⎬cos(θ ) + ⎨ ⎬sin (θ )⎥
⎩c⎭
⎦
⎣⎩ c ⎭
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)
If a = b = 1 then c = √2
A = 45o
a cos (θ) + b sin(θ) = c{cos(A – θ)}
cos (θ) + sin(θ) = √2 {cos(45o – θ)}
If we started with the expression b cos (θ) - a sin(θ)
⎤
⎡⎧ b ⎫
⎧a ⎫
b cos (θ) - a sin(θ) = c ⎢⎨ ⎬cos(θ ) − ⎨ ⎬sin (θ )⎥
⎩c ⎭
⎦
⎣⎩ c ⎭
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)}
o
If a = b = 1 then c = √2
A = 45
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
3cos( θ) − 4sin( θ) = 5 cos(θ + 53.13o )
Check put θ = 45o 3cos( θ) − 4sin( θ) = −0.707 and 5 cos(θ + 53.13o ) = -0.707
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 5 sin(θ + 143.13o ) = -0.707
©D.J.Dunn www.freestudy.co.uk 4
WORKED EXAMPLE No.2
sin (2θ )
= tan (θ )
1 + cos(2θ )
SOLUTION
Show that
Substitute sin(2θ) = 2sin(θ) cos(θ)
Substitute cos(2θ) = 2cos2(θ) – 1
Simplify and
sin (2θ )
2sin (θ )cos(θ )
=
1 + cos(2θ )
1 + cos(2θ )
sin (2θ )
2sin (θ )cos(θ )
2sin (θ )cos(θ )
=
=
2
1 + cos(2θ ) 1 + 2 cos (θ ) − 1
2 cos 2 (θ )
sin (2θ )
sin (θ )
=
= tan (θ )
1 + cos(2θ ) cos(θ )
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
- 2cos2(θ)} + 3cos(θ) – 3/2 = 0
2 - 2cos2(θ)} + 3cos(θ) = 7/2
2cos2(θ) - 3cos(θ) - 3/2 = 0
3 ± 32 + (4)(2)(3 / 2) 3 ± 9 + 12 3 ± 21
=
=
(2)(2)
4
4
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
cos( θ) =
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
P = {200 sin(2πft)} {5 sin(2πft)} = 1000 sin2(2πft)
1 − cos(2A)
Use the double angle formula sin 2 (A) =
2
⎛ 1 − cos{2(2 π f t )} ⎞
P = 1000⎜
⎟ = 500 − 500cos{4 π f t}
2
⎠
⎝
The frequency is doubled and P varies 0 to 1000
(see the plot)
©D.J.Dunn www.freestudy.co.uk 5
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)
cosh 2 ( θ) + sinh 2 ( θ) = 1
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) m sin(A) sin(B)
cosh(x) + sinh(x) = ex
cosh(x) - sinh(x) = e-x
dy
= cosh(x)
dx
dy
If y = cosh(x) then
= sinh(x)
dx
dy
= cos(x)
dx
dy
If y = cos(x) then
= −sin(x)
dx
∫ sinh(x)dx = cosh(x) + C
∫ sin(x)dx = - cosh(x) + C
∫ cosh(x)dx = sinh(x) + C
∫ cos(x)dx = sin(x) + C
y = sinh(x) then
©D.J.Dunn www.freestudy.co.uk 6
y = sin(x) then
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 v 2 = Rg
{µ + tanθ}
{1 − µtanθ}
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
(Answer 45o)
4.
Prove the following by using standard trigonometric identities.
i.
Solve the smallest positive value of θ
1 − cos(2θ )
1 − cos(2θ )
= tan (θ ) ii.
= tan 2 (θ )
sin(2 θ)
1 + cos(2 θ)
tanA + tanB
1 − tanA tanB
5.
Show that tan (A + B) =
6.
Change 5cos(α) + 2 sin(α) into cosine form
(Answer 5.385 cos(α -21.8o)
7.
Change 3 sin( θ) − cos( θ) 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:
σx + σy σx − σy
σθ =
+
cos(2 θ) + τsin(2θ)
2
2
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)
©D.J.Dunn www.freestudy.co.uk 7