Chapter 7
CHAPTER 6 SOLUTIONS TO Reinforcement exercises IN TRIGONOMETRY
6.3.1 Radian measure and the circle
6.3.1A.
Express as radians
i) 36o ii) 101o iii) 120o iv) 250o
v) 340o vi) – 45o vii) – 110o viii) 15o
ix) 27o x) 273o
Solution
All we have to remember in converting between degrees and radians is that
2π radians = 360 degrees
Then:
i) 36o = = radians = radians
Note that it is usual to leave radian measure in such form, rather than convert it to decimal form. Once you get the hang of this sort of problem it is little more than practice in cancelling down fractions.
ii) Since 2π radians = 360 degrees, 1o = radians and so 101o = radians
iii) 120o = radians = radians
iv) 250o = = radians
v) 340o = = radians
vi) – 45o = – = – radians
vii) – 110o = – = – radians
viii) 15o = = radians
ix) 27o = = radians
x) 273o= = radians
6.3.1B.
Express the following radian measures in degrees in the range 0 ≤ θ < 360o.
i) ii) 14( iii) – iv)
v) vi) vii) viii)
ix) – x)
Solution
Whereas to convert degrees to radians we have to multiply by to go in the opposite direction and convert radians to degrees we have to multiply by .
i) radians = ( = 120o
ii) 14( radians = 14( ( = 7 ( 360o = 360o on reduction to the 360 range
iii) – radians = – ( = – 90o = 270o
iv) radians = ( = 60o
v) radians = ( = 30o
vi) radians = ( = 450o = 90o on reduction to 360 range
vii) radians = ( = 40o
viii) radians = ( = 225o
ix) – radians = – ( = – 72o = 308o
x) radians = ( = 15o
6.3.1C.
Determine the length of arc and the area of the sector subtended by the following angles in a circle of radius 4cm.
i) 15o ii) 30o iii) 45o iv) 60o
v) 90o vi) 120o vii) 160o viii) 180o
Solution
The length of arc of a circle of radius r subtended by an angle θ RADIANS at the centre is given by s = rθ, and the corresponding area of sector is A = r2θ. since the radius is the same, 4cm in each case, we can write these as
s = 4θ and A = 8θ
and the calculations are then just an exercise in changing degrees to radians.
i) For θ ’ 15o = radians we have s = 4 = cm and A = 8 = cm2
ii) For θ ’ 30o = radians we have s = 4 = cm and A = 8 = cm2
iii) For θ ’ 45o = radians we have s = 4 = π cm and A = 8 = 2π cm2
iv) For θ ’ 60o = radians we have s = 4 = cm and A = 8 = cm2
v) For θ ’ 90o = radians we have s = 4 = 2π cm and A = 8 = 4π cm2
vi) For θ ’ 120o = radians we have s = 4 = cm and A = 8 = cm2
vii) For θ ’ 160o = radians we have s = 4= cm and A = 8 = cm2
viii) For θ ’ 180o = π radians we have s = 4π cm and A = 8π cm2
6.3.2 Definitions of the trig ratios
6.3.2A.
Write down the exact values of sine, tan, sec, and complementary ratios for all 'special' angles 0, π/2, π/3, π/4, π/6.
Solution
This may seem like a lot to remember, but with a few aids it is not too bad. It is also very useful to be able to recall these results easily (particularly those for sine and cosine), as they occur frequently (pun intended!) in such topics as AC circuit theory and signal analysis. All you really need to remember are the 30-60 and 45 triangles given in Figure 6.4, and then all of the results are just a matter of definition. Try to get used to using the surds, such as and (See Section 1.2.7). It is much better to leave them as they are until the end of calculation, rather than immediately replacing them by decimal approximations. All you have to remember is, for example ()2 = 2, which is something like how we handle the imaginary j in complex numbers (Chapter 12), by using j2 = ( 1 whenever possible. If you didn’t get all the results right first time, use Figure 6.4 to check the following answers.
| |θ |0 |/2 |/3 |/4 |/6 |
| |sin θ |0 |1 | | | |
| |tan θ |0 |nd | |1 | |
| |sec θ |1 |nd |2 | | |
| |cos θ |1 |0 | | | |
| |cot θ |nd |0 | |1 | |
| |cosec θ |nd |1 | | |2 |
(nd means not defined, because in principle we are trying to divide by zero – in practice, however, it is common to regard, for example, tan(/2) as ‘infinity’).
6.3.2B.
Classify as odd or even functions: cos, sin, tan, sec, cosec, cot.
Solution
Again, you either know or don't know that cosine is even and sine is odd. So tan = sin/cos is odd. sec = 1/cos is even. cosec = 1/sin is odd. cot = cos/sin is odd. Remember such results as Odd ( Even = Odd, O/E = O, O ( O = E, etc.
6.3.2C.
Express as trig functions of x (n is an integer) :-
i) sin(x + n() ii) cos(x + n() iii) tan(x + n()
iv) sin v) cos vi) tan
where n is an integer.
Solution
There is a bit more meat to this question! The safest approach is to use the corresponding compound angle formulae. Also note that we will be frequently using such results as cos n( = (( 1)n, sin n( = 0, etc, so make sure that you understand these results.
i) From the compound angle formula for sin(A + B) we have
sin(x + n() = sin x cos nπ + cos x sinnπ
= sin x cos nπ = (– 1)n sin x
ii) From the compound angle formula for cos(A + B) we have
cos(x + n() = cos x cos nπ + sin x sinnπ
= cos x cos nπ = (– 1)n cos x
iii) tan(x + n() = = = tan x
iv) sin= sin x cos + sin cos x
The form of this depends on whether n is odd or even. the easiest way to find the general result is to try a few specific values of n and then see what the general pattern is. We then find that:
If n is odd then cos = 0 and sin = (– 1)(n–1)/2
So sin= (– 1)(n–1)/2 cos x if n is odd.
On the other hand if n is even then sin = 0 and cos = (– 1)n/2
So sin= (– 1)n/2 sin x if n is even.
v) cos= cos x cos – sin sin x
= (– 1)n/2 cos x if n is even.
– (– 1)(n–1)/2 sin x if n is odd
= (– 1)(n +1)/2 sin x if n is odd.
vi) tan=
Again we get different answers depending on whether n is odd or even. Using the results of iv) and v) we get
If n is even:
tan= = tan x
and if n is odd
tan= = – cot x
6.3.3 Sine and cosine rules and the solution of triangles
With the standard notation solve the following triangles.
i) A = 70o , C = 60o , b = 6
ii) B = 40o , b = 8, c = 10
iii) A = 40o , a = 5, c = 2
iv) a = 5 b = 6, c = 7
v) A = 40o , b = 5, c = 6
vi) A = 120o , b = 3, c = 5
Solution
i) In this case we are given two angles, A = 70o and C = 60o, and one side, b = 6. From A = 70o and C = 60o we know that the third angle must be B = 180o – 70o – 60o = 50o.
We can now use the sine rule to find the two other sides:
= =
so
= =
Hence
a = = 7.3601 to 4dp
c = = 6.7831 to 4dp
ii) This is the case of two sides b = 8 and c = 10, and a non-included angle B = 40o. In this case we can get two possible triangles to fit the information given - it is the 'ambiguous case'. Firstly, apply the sine rule to attempt to find 'sin C'. We have
= = =
so
sin C = = 0.8035 to 4dp
Now although this looks like one answer, we have to remember that
sin (180o – C) = sin C
So if we take C = C1 to be the acute angle solution, then there is another solution with an obtuse angle C2 = 180o – C1. Both solutions will produce triangles satisfying the given condition. For the acute angle result we find
C1 = 53.46o to 2 dp
The obtuse angle result is then
C2 = 180o – 53.46o = 126.54o to 2 dp
So we have two possible triangles for which to find A and a.
C1 = 53.46o
Then
A = 180o – 53.46o – 40o = 180o – 53.46o = 86.54o
and
=
or
a = = 12.42 to 2dp
C2 = 126.54o
Then
A = 180o – 126.54o – 40o = 13.46o
Note: If it had happened that the larger value of C produced a negative answer here then the answer would not be a valid solution - we would effectively only have one triangle.
So
=
or
a = = 2.9 to 2dp
iii) For A = 40o, a = 5, c = 2 we again have two sides and a non-included angle. The sine rule gives
=
so
sin C = = 0.2571 to 4dp
This gives two values of C, as before:
C1 = 14.9o to 2 dp
C2 = 180o – 14.9o = 165.1o to 2 dp
In the acute case we obtain for B:
B = 180o – A – C1 = 180o – 40o – 14.9o = 125.1o
But in the obtuse case we see that A and C2 add to give 165.1o + 40o = 205.1o. Since this is greater than 180o no triangle exists for this value of C which must therefore be rejected.
For the allowed value of C = 14.9o we find the remaining side from
=
or
b = = 6.36 to 2dp
iv) a = 5, b = 6, c = 7 is the case of all three sides and finding the angles. We can use the cosine rule to find any of the angles.
From a2 = b2 + c2 – 2bc cos A we have
cos A = = = 0.7143
and
A = 44.42o
Similarly for B we find
cos B = = 0.543
and so
B = 57.11o
We can then find C from 180o – A – B = 180o – 44.42o – 57.11o = 78.47o
v) A = 40o, b = 5, c = 6 is the case of two sides and the included angle
We have
a2 = b2 + c2 – 2bc cos A = 52 + 62 – 2(5)(6) cos 40o
= 25 + 36 – 60 cos 40o = 15.04373
and so
a = 3.88
Now the sine rule gives
= =
so
sin C =
from which
C = 84.02o
Similarly we get for B:
B = sin– 1 = 55.98o
It pays to check our answer by confirming that the angles add up to 180o (or alternatively of course we could forego the check and determine, say, B by subtraction from 180o). So, the final answer is
a = 3.88, B = 55.98o, C = 84.02o
vi) A = 120o, b = 3, c = 5 is as for v) but now we have an obtuse angle – but the calculations are basically the same, just remembering that cos 120o = – .
We have
a2 = 32 + 52 – 2(3)(5) cos 120o
= 9 + 25 + 30 (0.5) = 49
and so
a = 7
Now the sine rule gives
= =
So
sin B =
from which
B = 21.77o
Similarly we get for C:
C = sin– 1 = 38.21o
So the answer is
a = 7, B = 21.77o, C = 38.21o
6.3.4 Graphs of trigonometric functions
Sketch the graphs of
i) 2 sin (2t + ) ii) 3 cos (3t – )
Solution
Both sin and cosine are of course just continuous waves, and in sketching them we only have to size and locate them. In the general form x = A sin(or cos) ((t + () A gives the amplitude, or the ‘height’ of the wave above the mean level, ( determines the number of oscillations in a given period of t and ( fixes the location of the graph by ensuring that at t = 0, x = A sin (, ie it passes through A sin ( on the vertical axis, with t plotted horizontally. Using these ideas you should obtain the graphs in the figures.
[pic]
6.3.5 Inverse trigonometric functions
6.3.5A.
Find the principal value and general solutions for sin–1x and cos–1x for the following values.
i) 0 ii) iii) –
iv) v) vi) –
Solution
The inverse sine and cosine can be evaluated from the general results (given in radian form)
sin–1x = nπ + (–1)n PV
cos–1 x = 2nπ ± PV
where PV denotes the corresponding principle values, which take the ranges
– [pic] sin ( 1 x [pic]
0 [pic] cos ( 1 x [pic] (
Applying these results, with the appropriate principal values we obtain the table below
| x |sin–1x |sin–1x |cos–1x |cos–1x |
| |PV |GS |PV |GS |
|i) 0 |0 |n( | |2nπ ± |
|ii) | |n( + (–1)n | |2nπ ± |
|iii) – |– |n( +(–1)n+1 | |2nπ ± |
|iv) | |n( + (–1)n | |2nπ ± |
|v) | |n( + (–1)n | |2nπ ± |
|vi) – |– |n( +(–1)n+1 | |2nπ ± |
where n is an integer
6.3.5B.
Find the principal value and general solutions for tan–1x for the following values of x
i) 0 ii) 1 iii) iv) –1 v) –
Solution
Using tan–1 x = nπ ± PV where the PV is in the range ( [pic] tan ( 1 x [pic] we obtain the following results.
|tan–1x |0 |1 | |– 1 |– |
|PV |0 | | |– |– |
|GS |nπ |nπ + |nπ + |nπ – |nπ – |
6.3.6 The Pythagorean identities – 'cos2 + sin2 = 1'
6.3.6A.
i) For c = cos θ, simplify
a) b) c)
ii) For s = sin θ, simplify
a) b) c)
iii) For t = tan θ, simplify
a) b) c)
Solution
All these questions are simply variations on the Pythagorean identities:
cos2θ + sin2θ ∫ 1
1 + tan2θ ∫ sec2θ
cot2θ + 1 ∫ cosec2θ
i) For c = cos θ we have
a) = = = sinθ
b) = = = cot θ
c) = = = tan2θ
ii) For s = sin θ we have
a) = = = cos θ
b) =
c) = = = sec θ tan θ
iii) For t = tan θ we have
a) = = = sec θ
b) = = = sin θ cos θ
c) = = = cos θ cot θ
6.3.6B.
Eliminate θ from the equations
i) x = a cos θ, y = b sin θ ii) x = a sin θ, y = b tan θ
Solution
i) With x = a cos θ, y = b sin θ we have = cos θ and = sin θ and so using the Pythagorean identity:
cos2 θ + sin2 θ ’ 2 + 2 = 1
ii) With x = a sin θ, y = b tan θ we have
y = b tan θ = b = b
But sin θ = so
y = b = =
6.3.6C.
For the following values of sin θ, find the corresponding values of cos θ and tan θ without using your calculator, giving your answers in surd form, and assuming that θ is acute.
i) ii) iii) – iv)
Solution
i) If sin θ = then
cos θ = = = = =
So
tan θ =
ii) If sin θ = then
cos θ = = = = = =
So
tan θ =
iii) If sin θ = then
cos θ = = = = = =
So
tan θ =
6.3.6D.
If r cos θ = 3 and r sin θ = 4 determine the positive value of r, and the principal value of θ.
Solution
Square and add to get
(r cos θ)2 + (r sin θ)2 = r2 cos2 θ + r2 sin2 θ = r2 (cos2 θ + sin2 θ)
= r2 = 32 + 42 = 9 + 16 = 25
so r = 5 (r is always taken positive)
From r cos θ = 3 we therefore have 5 cos θ = 3 or cos θ = from which
θ = cos–1 = 53.13∞ to 2dp
6.3.7 Compound angle formulae
6.3.7A.
Prove the following
i) sin 3θ = 3 sin θ – 4 sin3θ ii) cos 3θ = 4cos3θ – 3 cosθ
iii) = cos θ – sin θ iv) cot θ – tan θ = 2 cot 2θ
v) cot 2θ =
Solution
i) By the compound angle formula we have
sin 3θ = sin (2θ + θ) = sin 2θ cos θ + sin θ cos 2θ
Using the double angle formula on the sin 2θ and cos 2θ gives
sin 3θ ’ 2sin θ cos θ cos θ + sin θ ( cos2θ – sin2θ)
= 3 sin θ cos2θ – sin3θ = 3 sin θ (1 – sin2θ) – sin3θ
= 3 sin θ – 4 sin3θ
ii) By the compound angle formula we have
cos 3θ = cos (2θ + θ) = cos 2θ cos θ – sin θ sin 2θ
Using the double angle formula on the sin 2θ and cos 2θ gives
sin 3θ ’ ( cos2θ – sin2θ)cos θ – sin θ 2sin θ cos θ
= cos3θ – sin2θ cos θ – 2 (1 – cos2θ) cosθ
cos3θ – (1 – cos 2θ) cos θ – 2 (1 – cos2θ) cosθ
= 4cos3θ – 3 cosθ
iii) =
= = cos θ – sin θ
iv) cot θ – tan θ = –
= = 2
on cross mulitplying and using double angle formulae
= 2 cot 2θ
v) cot 2θ =
6.3.7B.
Without using a calculator or tables evaluate
i) sin 15( cos 15( ii) sin 15( iii) tan(π/12) iv) cos(11π/12)
v) tan(7π/12) vi) cos 75(
Solution
In this and the next question you need to be adept at dealing with surds, so if you are a bit rusty on this, go back to Section 1.2.7 for a bit of revision.
i) sin 15( cos 15( = 2 sin 15( cos 15( = sin[ 2 (15() ] = sin 30(
= . =
ii) sin 15( = sin (45( – 30() = sin 45( cos 30( – sin 30( cos 45(
= – =
=
iii) tan(π/12) = tan 15( = tan (45( – 30() =
= = = =
= = 2 –
iv) cos(11π/12) = cos 165( = cos (120( + 45()
= cos 120( cos 45( – sin 120( sin 45(
= – – = – = –
v) tan(7π/12) = tan = =
= = = – = – (2 + )
vi) cos 75( = cos (30( + 45() = cos 45( cos 30( – sin 30( sin 45(
= – =
which of course is just sin (90( – 75() = sin 15( as I'm sure you will have noticed!
6.3.7C.
Evaluate
i) sin 22.5o ii) cos 22.5o iii) tan 22.5o,
given that cos 45o = 1/.
Solution
This question uses the double angle formulae:
sin 2A ∫ sin A cos A
cos 2A ∫ cos2 – sin2 A
∫ 2 cos2A – 1
∫ 1 – 2 sin2A
tan A ’
i) sin 22.5o = sin (45o/2) = by the double angle formula for cos 2A
= =
=
ii) cos 22.5o = cos (45o/2) =
= =
=
iii) tan 22.5o = = from i) and ii)
= = =
6.3.7D.
Express the following products as sums or differences of sines and/or cosines of multiple angles
i) sin 2x cos 3x ii) sin x sin4x iii) cos 2x sin x iv) cos 4x cos 5x
Solution
These questions use the compound angle formulae 'backwards'.
i) We know that sin 2x cos 3x will occur in the expansion of sin (2x + 3x) and sin (2x – 3x). Specifically, we have
sin 5x = sin(2x + 3x) = sin 2x cos 3x + sin 3x cos 2x
sin (– x) = sin(2x – 3x) = sin 2x cos 3x – sin 3x cos 2x
Adding these gives
sin 5x + sin (– x) = sin 5x – sin x = 2 sin 2x cos 3x
on using the fact that sin x is odd.
So
sin 2x cos 3x =
ii) Remembering 'cos (+) = cos cos – sin sin' we know that sin x sin 4x occurs in cos ( x – 4x) = cos 3x, and in cos (x + 4x) = cos 5x. Thus
cos 5x = cos (x + 4x) = cos x cos 4x – sin x sin 4x
cos 3x = cos (x – 4x) = cos x cos 4x + sin x sin 4x
Subtracting and dividing by 2 then gives
sin x sin 4x =
iii) We have
sin 3x = sin (2x + x) = cos 2x sin x + sin 2x cos x
and
sin (– x) = – sin x = – sin (2x – x) = – sin 2x cos x + sin x cos 2x
from which
cos 2x sin x =
iv) The argument should now be familiar:
cos 9x = cos (4x + 5x) = cos 4x cos 5x – sin 4x sin 5x
cos x = cos (4x – 5x) = cos 4x cos 5x + sin 4x sin 5x
and addition gives
cos 4x cos 5x =
6.3.7E.
Prove the following identities (Hint: put P = (A + B)/2, Q = (A – B)/2 in the left-hand sides, expand and simplify and re-express in terms of P and Q)
i) sin P + sin Q ∫ 2sin cos
ii) sin P – sin Q ∫ 2cossin
iii) cos P + cos Q ∫ 2coscos
iv) cos P – cos Q ∫ –2sinsin
Solution
The question hints at the method of proofs of these relations, but there are some shortcuts for the alert. We prove the first in full.
i) Put A = and B = , or P = A + B and Q = A – B.
Then
sin P + sin Q = sin (A + B) + sin (A – B)
= sin A cos B + sin B cos A + sin A cos B – sin B cos A
= 2 sin A cos B
= 2 sin cos
ii) - iv) can be proved similarly, or we can use i) along with such ploys as replacing Q by – Q in i) to get ii) and, for example in iii)
iii) cos P + cos Q ∫ sin (90( – P) + sin (90( – Q)
= 2 sin cos
2 sin cos
= 2 cos cos
6.3.8 Trigonometric equations
6.3.8A.
Give the general solution to each of the equations:
i) cos θ = 0 ii) cos θ = –1 iii) cos θ = –
iv) sin θ = 0 v) sin θ = –1 vi) sin θ =
vii) tan θ = 0 viii) tan θ = –1 ix) tan θ =
Solution
These are really examples in inverse functions, and we have in fact done some of them already in RE 6.3.5. We only need to use the results of Section 6.2.5 for the general forms of the inverse trig functions:
sin–1x = nπ + (–1)n PV
cos–1 x = 2nπ ± PV
tan–1 x = nπ + PV
Where PV is the principal value, and n is an integer.
i) An equation of the form cos θ = a has the general solution θ = 2nπ ± PV, so cos θ = 0 has the general solution (GS)
θ = 2nπ ±
since the PV of cos–1 0 is .
ii) cos θ = –1 has the GS θ = 2nπ ± π = (2n ± 1)π
since the PV of cos–1 (– 1) is π.
iii) The PV of θ for cos θ = – is θ = π – = and so the GS is
θ = 2nπ ±
iv) The GS of sin θ = a is θ = nπ ± (– 1)n PV. Since the PV of θ for sin θ = 0 is θ = 0, the GS of sin θ = 0 is just
θ = nπ
v) The GS of sin θ = –1 is
θ = nπ ± (– 1)n
since the PV of sin–1 (– 1) is – .
vi) No, sin θ = is not a misprint - just a trap for the unwary! Since, for real angles, θ, – 1 [pic] sin θ [pic] 1 it follows that sin θ = has no solutions at all.
vii) The GS of tan θ = a is
θ = nπ + PV
and the PV of θ for tan θ = 0 is θ = 0, so the GS of tan θ = 0 is
θ = nπ
viii) The GS of tan θ = – 1 is
θ = nπ +
since the PV of tan–1 (– 1) is .
ix) For tan θ = the PV of tan–1 () is and so the GS is
θ = nπ +
6.3.8B.
Find the general solutions of the equations
i) cos 2θ = 1 ii) sin 2θ = sinθ
iii) cos 2θ + sinθ = 0 iv) cos 2θ + cos 3θ = 0
v) sec2θ = 3 tan θ – 1
Solution
i) The GS of cos 2θ = 1 for 2θ is, from the previous question
2θ = 2nπ
and so for θ the GS is
θ = nπ
ii) To solve sin 2θ = sinθ we must get everything in terms of simple sines and cosines, for which we use the double angle formula
sin 2θ = 2 sinθ cos θ = sinθ
This can be rewritten as
(2 cos θ – 1) sinθ = 0
which shows that
2 cos θ – 1 = 0 or sinθ = 0
ie either cos θ = or sin θ = 0.
The GS of sin θ = 0 is θ = nπ The GS of cos θ = is θ = 2nπ ± which together give the full set of solutions for θ.
An alternative approach to this sort of question is to use the GS of sin A = sin B given in Section 6.2.8, namely A = 2nπ + B or A = (2n + 1)π – B. So the GS of sin 2θ = sin θ will be
2θ = 2nπ + θ or (2n + 1)π – θ
Hence
θ = 2nπ or θ = where n is an integer
I will leave it as an exercise for you to confirm that this solution is equivalent to the one we obtained above (Evaluate both for n = 0, 1, 2, ...)
iii) To solve cos 2θ + sinθ = 0 we must again get everything in terms of simple sines and cosines, for which we use the double angle formula
cos 2θ = 1 – 2sin2θ
Then the equation can be rewritten as
1 – 2sin2 θ + sinθ = 0
or
2 sin2 θ – sin θ – 1 = 0
= (2 sin θ + 1)(sin θ – 1)
Hence either sin θ = – or sin θ = 1.
The GS of sin θ = 1 is θ = π. The GS of sin θ = is θ = nπ – (– 1)n . These results can be combined together in the general solution for θ:
θ = π
An alternative approach is to use the GS of cos A = cos B given in Section 6.2.8, after converting the sin θ to a complementary cosine.
iv) In cos 2θ + cos 3θ = 0 use of multiple angle formulae is going to be messy. It is easier to use the result
cos P + cos Q ∫ 2cos cos
We have
cos 2θ + cos 3θ = 2 cos cos = 0
So
cos = 0 or cos = 0
Hence
= π and therefore θ = π
or
= π and therefore θ = (2n + 1) π
These two apparently different looking results can be combined by noting that when n = 5m + 2 where m is any integer (and of course n can be any integer)
π = π = (2m + 1) π
So the GS is
θ = π where n is any integer
v) sec2θ = 1 + tan2θ = 3 tan θ – 1 and so
tan2 θ – 3 tan θ + 2 = 0
Factorising this quadratic in tan θ gives
(tan θ – 1)(tan θ – 2) = 0
So
tan θ = 1 or tan θ = 2
Therefore
θ = or θ = 1.11 radians for the PV
Hence the GS is
θ = nπ + , nπ + 1.11
6.3.9 The a cos θ + b sin θ form
6.3.9A.
Write the following in the form a) r sinb) rcos
i) sin θ – cos θ ii) cos θ + sin θ iii) sinθ – cos θ
iv) 3 cos θ + 4 sin θ
Solution
i) a) We start by writing
sin θ – cos θ = r sin
= r sin θ cos α + r cos θ sin α
So
r cos α = 1 and r sin α = – 1
Squaring and adding to eliminate α gives r = , and then
cos α = and sin α = –
The principal value of α is then
α = – 45( = –
So we now have
sin θ – cos θ = = sin
b) A little more quickly now:
sin θ – cos θ = r cos
= r cosθ cos α – r sinθ sin α
So
r sin α = – 1 and r cos α = 1
and again r = , so
sin α = – and cos α = –
The principal value of α is then
α = – 135( = –
So
sin θ – cos θ = cos
ii) a) cos θ + sin θ = r sin
= r sinθ cos α + r cosθ sin α
So
r sin α = and r cos α = 1
from which r = 2 and then
cos α = and sin α =
The principal value of α is then
α = 60( =
So we have
cos θ + sin θ = 2 sin
b) cos θ + sin θ = r cos
= r cos θ cos α – r sin θ sin α
So
r cos α = and r sin α = (1
from which r = 2 and then
sin α = – and cos α =
The principal value of α is therefore
α = – 30( = –
So we have
cos θ + sin θ = 2 cos
iii) a) sin θ – cos θ = r sin
= r sin θ cos α + r cos θ sin α
So
r cos α = and r sin α = – 1
from which r = 2 and then
sin α = – and cos α =
The principal value of α is then
α = –
So we have
sin θ – cos θ = 2 sin
b) sin θ – cos θ = r cos
= r cos θ cos α – r sin θ sin α
So
r sin α = – and r cos α = – 1
from which r = 2 and then
cos α = – and sin α = –
The principal value of α is therefore
α = – 150( = –
So we have
sin θ – cos θ = 2 cos
iv) a) 3 cos θ + 4 sin θ = r sin
= r sinθ cos α + r cosθ sin α
So
r sin α = 3 and r cos α = 4
From which r = 5, and then
cos α = and sin α =
The principal value of α is then
α = 36.87( = 0.6435 radians
So we now have
3 cos θ + 4 sin θ = 5 sin)
b) 3 cos θ + 4 sin θ = r cos
= r cosθ cos α – r sinθ sin α
So
r sin α = – 4 and r cos α = 3
and again r = 5, so
cos α = and sin α = –
The principal value of α is then
α = – 53.13(
So
3 cos θ + 4 cos θ = 5 cos
6.3.9B.
Determine the maximum and minimum values of each of the expressions in Question A, stating the values where they occur, in the range 0 [pic] θ [pic] 2π.
Solution
i) a) sin θ – cos θ = sinfrom A.
This has a maximum value of when θ – = , or
θ =
It has a minimum value of – at θ – = , or
θ =
So there is a maximum of at θ = and a minimum value of – at θ =
ii) cos θ + sin θ = 2 sin
This has a maximum of 2 at θ + = , ie θ =
It has a minimum of – 2 at θ + = , ie θ =
iii) sin θ – cos θ = 2 sin
This has a maximum of 2 at θ – = , ie θ =
It has a minimum of – 2 at θ – = , ie θ =
iv) 3 cos θ + sin θ = 5 sin
This has a maximum of 5 at θ + 0.64 = , ie θ = 0.93 radians
It has a minimum of – 5 at θ + 0.64 = , ie θ = 4.07 radians
6.3.9C.
Find the solutions, in the range 0 [pic] θ [pic] 2π, of the equations obtained by equating the expressions in Question A to a) 1 b) –1.
Solution
i) a) sin θ – cos θ = sin= 1 from A.
So
sin=
In the range 0 [pic] θ [pic] 2π the solutions are
θ – = ,
and therefore
θ = ,
b) sin θ – cos θ = sin= – 1, gives
sin= –
In the range 0 [pic] θ [pic] 2π the solutions are
θ – = – ,
and therefore
θ = 0, π
ii) a) cos θ + sin θ = 2 sin= 1
So
sin=
In the range 0 [pic] θ [pic] 2π the solutions are
θ + = ,
and therefore
θ = ,
b) cos θ + sin θ = 2 sin= – 1
So
sin= –
In the range 0 [pic] θ [pic] 2π the solutions are
θ + = ,
and therefore
θ = ,
iii) a) sin θ – cos θ = 2 sin= 1
So
sin=
In the range 0 [pic] θ [pic] 2π the solutions are
θ – = ,
and therefore
θ = , π
b) sin θ – cos θ = 2 sin= – 1
So
sin= –
In the range 0 [pic] θ [pic] 2π the solutions are
θ – = ,
and therefore
θ = , 2π
iv) a) 3 cos θ + 4 sin θ = 5 sin(θ + 0.64) = 1
So
sin(θ + 0.64) = = 0.2
In the range 0 [pic] θ [pic] 2π the solutions are, to two decimal places
θ + 0.64 = π – 0.2 = 2.9 and 2π + 0.2 = 6.48
and therefore
θ = 2.3, 5.84 radians
b) 3 cos θ + 4 sin θ = 5 sin= – 1
So
sin= – = – 0.2
In the range 0 [pic] θ [pic] 2π the solutions are
θ + 0.64 = 3.34 and 6.28
and therefore
θ = 2.7, 5.44
-----------------------
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
[pic]
................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.
Related searches
- chapter 7 learning psychology quizlet
- chapter 7 financial management course
- chapter 7 connect
- chapter 7 connect finance
- chapter 7 photosynthesis quizlet
- chapter 7 psychology quizlet
- psychology chapter 7 quiz quizlet
- chapter 7 psychology quizlet test
- chapter 7 learning quizlet
- psychology chapter 7 quizlet exam
- psychology chapter 7 learning
- psychology chapter 7 test questions