Trigonometry

Trigonometry



Problems and Solutions



with your fx-991ES or fx-


115ES Calculator












Dr Allen Brown








Cambridge 
Paperbacks

Cambridge Paperbacks

www.CambridgePaperbacks.com

First published by Cambridge Paperbacks 2018



© Allen Brown 2018

All rights reserved. No part of this publication may be reproduced or
transmitted in any form or by any means, electronic or mechanical, including
photocopy, recording, or any information storage and retrieval system without
permission in writing from the author.



Disclaimer

Although the author and publisher have made every effort to ensure that the
information in this book was correct during preparation and printing, the
author and publisher hereby disclaim any liability to any party for any errors
or omissions.

Read this First


Trigonometry has been a part of human knowledge

for many thousands of years and forms an important
part of a students skill set. Making the assumption you

are either curious about trigonometry or are pursuing
a maths course which has a strong trigonometry

content. In either case you should find the examples

in this ebook interesting.

Learning to do trigonometry using your fx calculator is

part of the learning objective of this ebook. Working

through each example will certainly strengthen your
knowledge of the subject.


As in other ebooks in this series, in every example the
keystrokes for your calculator are given. You are

therefore encouraged to repeat the same keystrokes

so to advance your understanding of trigonometry.
This ebook will convince you that your fx calculator

should be a constant companion as you work on

trigonometry problems wherever course you are
following.



Dr Allen Brown

Cambridgeshire

Contents

Example 1: ............................................................................ 2

Example 2: ............................................................................ 5

Example 3: ............................................................................ 7

Example 4: ............................................................................ 9

Example 5: .......................................................................... 12

Example 6: .......................................................................... 15

Example 7: .......................................................................... 18

Example 8: .......................................................................... 21
Example 9: .......................................................................... 25

Example 10: ........................................................................ 28

Example 11: ........................................................................ 33

Example 12 ......................................................................... 38

Example 13 ......................................................................... 41

Example 14 ......................................................................... 44

Example 15 ......................................................................... 48

Example 16 ......................................................................... 50








1

Example 1:

When you have trigonometric expressions of the kind,


α sin( ) + cos( ) 1


Then it’s possible to write it as

α sin( ) + cos( ) = cos( − ) 2


This comes from the identity,


cos( − ) = cos( ) cos( ) + sin( ) sin( )

When the right hand side of Eq:2 is expanded,


cos( − ) = cos( ) cos( ) + sin( ) sin( )

= [ sin( )] sin( ) + [ cos( )] cos( )


Compare with Eq:1,


= sin( ) and = cos( ) 3

When these are divided,



tan( ) = and γ = tan −1 ( ) 4


When you square and add,


2
2
= √ + 5


2

Express these in terms of your calculator notation,

then let = and = . Make sure your calculator
is in DEG mode (q(SETUP)3). The keystrokes for

Eq:4 and Eq:5 combined are,

q?Q(A)aQ(B)$)Q(:)sJ

(A)d+J(B)d


1. Express the following in R notation


4 sin( ) + 3 cos( )

r


A? 4=

B? 3=
x =








Leaving,

′
4 sin( ) + 3 cos( ) = 5 cos( − 53 7 . 37")

2. Express the following in R notation,


3.68 sin( ) + 6.49 cos( )

r

3

A? 3.68=
B? 6.49=


x =









3.68 sin( ) + 6.49 cos( )

′

= 7.46 cos( − 29 33 . 74")

3. Express the following in R notation,


22.56 sin( ) + 2.82 cos( )

r


A? 22.56=
B? 2.82=


x =








22.56 sin( ) + 2.82 cos( )


′
= 22.735 cos( − 82 52 29.94")
4

Example 2:


1. Prove that

sin(4 )
4 sin( ) cos( ) =
cos(2 )
Use the identities,

sin(2 ) = 2 sin( ) cos( )

sin(4 ) = 2 sin(2 ) cos(2 )


4 sin( ) cos( ) = 2sin(2 )


sin(4 )
2 sin(2 ) =
cos(2 )

which completes the identity.


2. Find the value of x (0 ≤ ≤ 2 ) which satisfies the

condition

sin( ) + sin ( ) = 0
2

Use the identity,

sin( ) = 2 sin ( ) cos ( )
2 2
therefore

2 sin ( ) cos ( ) + sin ( ) = 0
2 2 2

5

Write this as,

sin ( ) [2 cos ( ) + 1] = 0
2 2

This becomes
1
cos ( ) = −
2 2
1 1
= cos −1 (− ) or = 2 cos −1 (− )
2 2 2
The keystrokes for this are,
C2q>z1R2$)=







This is now M, test the result with the original

equation,

sin( ) + sin ( ) = 0
2
The keystrokes are,
jM)+jMa2$)=






Proving the answer is correct. The calculator is very

useful for confirming you have the right answer.


6

Example 3:

From the diagram on the left

derive an expression for the length

X.

Let the base side of the right angle

triangle be Y, then
tan( ) =

If the other angles in the isosceles
are Z, then

sin( ) sin( )
=


However + 2 = 180 ⇒ = 90 − ,
2

sin (90 − ) = sin(90) cos ( ) − cos(90) sin ( )
2 2 2


= cos ( )
2
Leaving,

sin( ) 1
= cos ( )
2
Inverting,


=
sin( )
cos ( )
2
7

or
sin( )
=
cos ( )
2
Finally,

sin( )
= tan( )

cos ( )
2
The keystrokes for your fx calculator for performing
this calculation are,

CQ(C)jQ(D))lQ(B))a

kJ(D)a2$)

o
In the diagram on page 7, C = 10.84 cm, D = 70.7 ,
o
B = 32.3 calculate the length X.
r

C? 10.84=

D? 70.7=

B? 32.3=








The length of X is 7.82 cm.



8

Example 4:



















A pattern is made from a disc of diameter F residing
on a rectangle of length A and width E. Derive an

expression for the length of the perimeter of the

pattern.


Mark out the chord on the
circle. The radius of the circle

is F/2 and the chord length x is

given by,


= sin ( ) 1
2

The arc length is given by where θ is measured in
2
radians. From the small triangle,

/2
cos ( ) = = 2
2 /2


9

2
2
sin ( ) = √1 − cos ( ) = √ 1 − ( )
2 2
The length x is,

2

√
2
2
= 1 − ( ) = √ − 3


From Eq:2,
−1
= cos ( ) 4
2
The top side has a length of,


( − ) +
2

2
2
= − √ − + cos −1 ( )


The perimeter of the pattern is therefore,

2
2
= 2 { + − √ − + cos −1 ( )}


The keystrokes for your fx calculator to perform the
calculation are,

C2(Q(E)+Q(A)psQ(F)dp

J(E)d$+J(F)q>J(E)aJ(F)
$))


10

In the pattern shown on page 9, the diameter of the

circle is F = 8.11 cm, the length of the rectangle is
A = 12.48 cm and its width E = 5.15 cm. What is the

perimeter of the pattern.


r

width E? 5.15=

length A? 12.48=
circle diameter F? 8.11=








The perimeter length of the pattern is 37.04 cm.


















11

Example 5:

Evaluate the following,



1 2
3 3
sin [cos −1 ( )] cos [tan −1 ( )]
4 5
3 4
5 6.482
tan [sin −1 (− )] sin [tan −1 ( )]
6 3
√11.48


1 The long way of performing calculations like this is

to let
3 3
= cos −1 ( ) or = cos( )
4 4


As you can see in the diagram
on the left,

4
√4 −3 2 √7
sin( ) = =
4 4
Therefore,

√7
= sin −1 ( )
4

For problem 1,


12

√7 √7
sin [sin −1 ( )] = = 0.661
4 4

The more efficient way of performing this calculation
is to use your fx calculator,

3
sin [cos −1 ( )]
4
The keystrokes for this are,
Cjq>3a4$))=









Likewise,

2
3
cos [tan −1 ( )]
5
Ckq?3a5$))=








3

5
tan [sin −1 (− )]
6

13

Clq<z5a6$))=











4
6.482
sin [tan −1 ( )]
3
√11.48
Cjq?6.482aS1

1.48$$))=











As you can see your fx calculator performs these
calculations with ease.











14

Example 6:

Derivative of an inverse trigonometric function. Let,

−1
= sin ( )
Then
−1
sin( ) = sin[sin ( )] =


Differentiating,

sin( ) =



cos( ) = 1

Dividing by cos(θ),

1 1
= =
cos( ) √1 − sin ( )
2
This becomes,
1
=
√1 − 2
Leaving,

−1 1
sin ( ) = 1
√1− 2


As you can see x < 1, to verify this result consider three
test values for = 0.49, 0.265, 0.857. Set calculator

15

to radian mode (q(SETUP)4). For the derivative

calculations, enter the following keystrokes,

For = −0.49
CqYq<Q(X)))$z0

.49=









For = 0.265
!oooooo0.265=








For = 0.857
!oooo857









Use the same values in the right hand side of Eq:1, use

the following keystrokes,
C1as1pQ(X)d

16

r


X? z0.49=







r


X? 0.265=








r

X? 0.857=







Which confirms the correctness of the result.


















17

Example 7:



From the diagram on the

left, derive an expression
for the length X in terms of

A, B and C. Also derive an

expression of the angle θ.


From the diagram


2
2
2
= + 1
2
2
2
= + ( + ) 2
From Eq:1
2
2
= √ −
From Eq:2
2
2
+ = √ −
Therefore,

2
2
2
2
√ − + = √ −
Leaving,

2
2
2
2
= √ − − √ − 3

18

From the diagram you will recognise,

sin(180 − ) =

Therefore,

180 − = sin −1 ( )


Leaving,

= 180 − sin −1 ( ) 4

Eq:3 and Eq:4 can be calculated at the same time on

your fx calculator by using the multi-stage feature;
separation by (:). The keystrokes are,



CsQ(C)dpQ(A)d$psQ(B)
dpJ(A)d$Q(:)

180pq<J(A)aJ(B)$)



In the diagram on page 18, A = 12.67 cm, B = 15.29 cm,
C = 18.41 cm. Calculate X and the angle θ.


r

C? 18.41=

A? 12.67=

B? 15.29=

19

=x








o
The length of X is 4.79 cm and the angle θ is 124 2
minutes and 24.3 seconds. In a similar diagram, A =
22.93 cm, B = 26.28 cm, C = 30.71 cm. Calculate X and

the angle θ.
r

C? 30.71=

A? 22.93=

B? 26.28=






=x








o
X is 7.58 cm and the angle is 119 ..
20

Example 8:




















A pattern is made by laying a triangle of length B and
base length A upon a circle whose diameter is D.

Derive an expression for the area of the pattern.


The area of the triangle is,

1

2

To calculate the area of the circle visible, consider the
diagram on the next page. Here the radius has been

marked out (length D/2) and the chord which has a

length A. The angle subtended by the two radii is θ,
and this related to D and A by,


/2
sin ( ) = = 1
2 /2



21

The area of the two blue triangles is given by,

2
2
1 √ − 2 √ − 2
2 × × × =
2 2 2 4

The area of the circle sector (the sum of the blue and
the red) is,

2
( )
2 2
However from Eq:1.

= sin −1 ( )
2
The sector area becomes



2
( ) sin −1 ( )
2
The red area is,



22

2
2 √ − 2
( ) sin −1 ( ) −
2 4

The area of the pattern is,

Area of green triangle + area of circle – red area.

2
1 2 2 √ − 2
+ ( ) − ( ) sin −1 ( ) +
2 2 2 4
Leaving,

2
√ − 2 2
= [ + ] + [ − sin −1 ( )]
2 2 4


The keystrokes for your fx calculator to perform this

calculation are (q(SETUP)4),
CQ(A)a2$(Q(B)+sQ(D)d

pJ(A)d$a2$)+J(D)da

$(qLpq<J(A)aJ(D)$
))

In the pattern on page 21 D = 15.49 cm, A = 11.04 cm

and B = 12.67 cm, calculate its area.
r

A? 11.04=

B? 12.67=

D? 15.49=

23

2
The area of the pattern is 240.78 cm . A similar
pattern has D = 20.22 cm, A = 14.71 cm and B = 11.78

cm, calculate its area.
r

A? 14.71=

B? 11.78=
D? 20.22=










2
The area of this pattern is 375.5 cm .













24

Example 9:

Find the solutions to equation between 0 ≤ ≤ 2 ,

sin(2 ) − cos( )
= 0
cos(2 ) + sin( ) − 1


Use the identities,

sin( + ) = sin( ) cos( ) + cos( ) sin( )
and

cos( + ) = cos( ) cos( ) − sin( ) sin( )
Then

sin(2 ) = sin( ) cos( ) + cos( ) sin( )
= 2 sin( ) cos( )



cos(2 ) = cos( ) cos( ) − sin( ) sin( )
2
2
= cos ( ) − sin ( )

Our equation becomes

2 sin( ) cos( ) − cos( )
= 0
2
2
cos ( ) − sin ( ) + sin( ) − 1

cos( ) [2 sin( ) − 1]
= 0
2
2
sin( ) − sin ( ) −[1 − cos ( )]



25

cos( ) [2 sin( ) − 1]
= 0
sin( ) − 2sin ( )
2

cos( ) [2 sin( ) − 1]
= 0
sin( ) [1 − 2 sin( )]

Leaving
cot( ) = 0

3
This occurs when cos(x) = 0 which is = and . We
2 2
can verify this result by substituting these values into
the original equation,


sin(2 ) − cos( )

cos(2 ) + sin( ) − 1


First set up your fx calculator in radian mode
(q(SETUP)4) then enter the following keystrokes,


Caj2Q(X))pkJ(X))Rk
2J(X))+jJ(X))p1


r

X? qLP2=








26

r


X? 3qLP2=








Which confirms the correctness of the results. A plot

of cot(x) is shown below, you can recognise where the
function passes through the x-axis in the region

0 ≤ ≤ 2 .

































27

Example 10:






































Here is a triangle on a grid, what is its area, the length
of its sides and the internal angles?


The coordinates of the apexes are,


A (-4, 5), B (5, -4), C (8, 6)

The area of the triangle can be determined by using a

determinant,



28

given by,

1
1 1 1
= | 2 2 1|
⊿
2
3 3 1

In this case,


1 −4 5 1
= | 5 −4 1|
⊿
2
8 6 1


To evaluate this determinant, use the MATRIX feature

on your fx calculator (w6). Then select, 11









Now enter the values from the determinant,

z4=5=1=5=z4=1=

8=6=1=C


All these numbers have gone into matrix A, now we
want to calculate the determinant of this matrix.

Enter the keystrokes,

0.5q(MATRIX)7q(MATRIX)3)=


29

2
The area of the triangle is therefore 58.5 units . To
calculate the side lengths between (X, Y) and (E, F),


2
2
√( − ) +( − )
The keystrokes for this calculation are (w1),

s(Q(X)pQ(E))d+(Q(Y)p

Q(F))d


The first side is A (-4, 5) to B (5, -4),
r

X? z4= E? 5=

Y? 5= F? z4=n








The second side A (-4, 5) to C (8, 6)


r

X? z4= E? 8=

Y? 5= F? 6=n

30

The third side B (5, -4) to C (8, 6)


r

X? 5= E? 8=
Y? z4= F? 6=n








The triangle lengths are,


AB (12.7279), AC (12.0416) and BC (10.4403)




Now use the cosine formula to calculate the angles in

the triangle,

2
2
2
2
2
2
+ − 2 + − 2 + − 2
cos −1 [ ] cos −1 [ ] cos −1 [ ]
2 2 2

The keystrokes for these three expressions, using a
multistage calculation are,




31

Cq>aQ(A)d+Q(B)dp

Q(C)dR2J(A)J(B)$)Q(:)



q>aJ(A)d+J(C)dpJ(B)
dR2J(A)J(C)$)Q(:)


q>aJ(B)d+J(C)dpJ(A)
dR2J(B)J(C)$)Q




To calculate the angles, A = 12.7279, B = 12.0416 and
C = 10.4403.


r
A? 12.7279=

B? 12.0416=

C? 10.4403=x


=x =x







o
When these are added you will get 180 as expected.





32

Example 11:

The diagram on the left shows a

2D vector rotated through an

angle θ. Rotation can be
represented by the action of a

2 × 2 matrix acting on a vector,

for example,

′ cos( ) −sin( )
( ) = ( ) ( )
′ sin( ) cos( )

The matrix operates on a vector to produce a rotated

vector. Consider the triangle
shown in the diagram on the

left. Rotate this triangle

o
through an angle of 27.5
around the origin. Start with

A we can see this has a

coordinate value of (2, 2),
therefore,


′ cos(27.5) −sin(27.5) 2
( ) = ( ) ( )
′ sin(27.5) cos(27.5) 2


This operation can be performed on your fx calculator,
first set up the Matrix A as a 2 × 2 with the keystrokes,

33

(w615C). Since we need to use 27.5 four

times in the calculation, make this the M value with
27.5=. Enter the following keystrokes,


q(MATRIX)21
kM)=zjM)=jM)=k

M)=








2
First calculation, set up matrix = ( ), use the
2
following keystrokes,

Cq(MATRIX)226

Enter the data, 2=2=






Perform the matrix multiplication,
Cq(MATRIX)3q(MATRIX)4








=


34

Which means,

2 0.8505
( ) → ( )
2 2.6975

3
Second calculation, set up matrix = ( ), use the
6
following keystrokes,
Cq(MATRIX)22

Enter the data, 3=6=









Perform the matrix multiplication,
Cq(MATRIX)3q(MATRIX)4=








Which means,
3 −0.109
( ) → ( )
6 6.7073



35

6
Third calculation, set up = ( ), use the following
0
keystrokes,
Cq(MATRIX)22

Enter the data, 6=0=










Perform the matrix multiplication,

Cq(MATRIX)3q(MATRIX)4=










Which means,

6 5.322
( ) → ( )
0 2.770


o
The rotated triangle through 27.5 anti-clockwise is
shown on the next page in red. You can see the point

A → A’, B → B’ and C → C’.



36

This operation is extremely important in computer
games. Every 3D graphics feature is made up of a huge

array of little polygons including triangles. To make

the feature rotate, every single little polygon must
rotate. This means an immense number of

calculations must be performed.

This is what graphics processors
are designed to do. The Asus

graphics card (on the left) hosts
the powerful GTX 750 graphics

processor.


37

Example 12























Referring to the diagram above, houses 2, 3 and 4 are
on the same latitude. Houses 1 and 3 are on the same

longitude. The separation of houses 2 and 4 is D km.
The angles between latitude and longitude for houses

2 and 3 are known, derive an expression for the

distance between houses 1 and 3.

Draw the triangle as shown

on the left. The known

angles are X and Y and the
distance house 1 to house 2

is A and the distance

between house 1 and house 4 is B. The angle at the
apex of the triangle is given by,

38

(90 − ) + (90 − ) = 180 − ( + )

The separation between houses 2 and 4 is D, we know,


sin[180 − ( + )] sin( )
=


Make B the subject of the equation,


sin( )
=
sin[180 − ( + )]

The distance between houses 1 and 3,


sin( )

therefore,


sin( ) sin( )
1−3 =
sin[180 − ( + )]


The keystrokes for performing this calculation are,

CaQ(D)jQ(X))jQ(Y))Rj

180pJ(X)pJ(Y))


Ensure your calculator is in degrees mode, keystrokes
(q(SETUP)3). The angle between houses 2 & 1

o
and is 35 and the angle between houses 4 & 1 is 40
degrees, D = 177 km. What is the distance between
houses 1 & 3?

39

r


D? 177=
X? 35=

Y? 40=








The distance between houses 1 & 3 is 67.56 km.

A mistake was made estimating the angle between

o
houses 2 & 1, it should have been 29.5 . What is the
corrected distance between houses 1 & 3?


r

D? 177=

X? 29.5=
Y? 40=








The corrected distance between houses 1 and 3 is

50.693 km.






40

Example 13
A linear function intercepts a cos(αx) function at the

following coordinates,


(−3.1, − 0.647), (−1.18, 0.047), (0.622, 0.692)

Determine the gradient and y-axis intercept and the

value of α.




Let the linear function be,


= +

Need to find the values α, b and c. From the first

coordinate,


−0.674 = −3.1 +

or,


3.1 − = 0.674 1

From the second coordinate,


0.047 = − 1.18 +


or

1.18 − = −0.047 2


41

Eq:1 and Eq:2 for simultaneous equations, use your fx

calculator to find the values of b and c, although in you
calculator they will be X and Y. Setup calculator into

equation mode (w5) and enter the values,








3.1=z1=0.674=

1.18=z1=p0.047=










=n =










The equation of the linear function is,


= 0.375 + 0.49

The value of α using the third coordinate,


0.692 = cos(0.622 )
42

0.622 = cos −1 (0.692)

1
= cos −1 (0.692)
0.622

The keystrokes for this calculation are,

w1q>0.692)a0

.622=









The value of α is therefore 1.296. A graph of the cos

function and linear function as shown here,




























43

Example 14
In this example we are going to create a parametric

graph by plotting


= cos( ) against = sin(2 )

So instead of x being regularly spaced, it becomes a

function in its own right. Create a table for both y(t)

and x(t). If you are using a fx-115ES calculator you can
create a table of both functions at the same time.



Set calculator into TABLE mode (w7). First

obtain the y(t) data. When you see f(X) in the display

enter the following keystrokes,

kQ(X))=

Start? z4=

End? 4=

Step? 0.3=












44

Create a table as shown below and transfer the data

from the calculator to the table.


y(t) -.65 -.84 -.96 -.99 -.94 -.8 -.58
y(t) -.32 -.03 0.27 0.54 0.76 0.92 0.99

y(t) 0.98 0.87 0.69 0.45 0.17 -.12 -.41
y(t) -.66 -.85 -.97 -.99 -.93 -.79


Now create the data for x(t), Cooo,

j2Q(X))=



45

Start? z4=

End? 4=
Step? 0.3=






































These are the values for x(t), transfer them to your

table which is on the next page.


46