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Game Math Chapter 5 Solutions
Game Math Chapter 5 Solutions
Frank Luna
www.gameinstitute.com
Saturday, January 8, 2004
1. Given the following right triangle information, find y and θ.
θ == . To solve for θ we take the inverse cosine of both sides
of the equation giving,
cos
xr
3 12
( ) ( )
( )
cos cos
−
1
()
θ
=
cos 3 12
−
1
.
θ
=
cos 3 12 75.522
−
1
≈ °
To solve for y we use the definition, sin
θ= , so ( )
yr
y =
12 sin 75.522
° ≈ .
11.618
2. Given the following right triangle information, find r and θ.
To solve for θ we use the definition, ( )
tan
θ = , and then apply the inverse tangent.
yx
So, with our numbers we have,
Frank Luna
Page 1
www.gameinstitute.com
We have, ()
736239423.007.png 736239423.008.png 736239423.009.png
Game Math Chapter 5 Solutions
() 5
tan
θ =
3
tan tan
−
1
( )
()
θ
=
tan
− 

3
θ
=
tan
− 
1 5

≈ °
59.036
.
3
Then, to solve for r , we employ the theorem of Pythagoras,
ab c
2
+ = , as follows,
2
35 r
2
+=
2
2
r =+= ≈ .
3 5
2
2
34 5.8309
Note you can also solve for r using the definition of either sine or cosine.
3. Given the following right triangle information, find x and θ.
We start with the definition of sine: ( )
sin
θ = . Substituting our numbers then gives,
yr
θ ==. To solve for θ we take the inverse sine of both sides of the
equation, which yields ( )
()
2 4 1 2
θ
=
sin 1 2 30
−
1
= ° , and recalling that ()
sin sin θ θ
1
( )
= , which
follows from definition. To solve for x we recall that ( )
cos
θ = , so
x r
xr θ
cos
() ( )
=
4 cos 30
° = = .
2 3 3.4641
4. Prove
1
+
cot
2
θ
=
csc
2
θ
.
We know () ()
sin
2
θ θ
+
cos
2
= . Dividing both sides of the equation by ()
1
sin θ gives:
2
Frank Luna
Page 2
www.gameinstitute.com
1 5
2
sin
−
=
736239423.010.png 736239423.001.png
Game Math Chapter 5 Solutions
sin
2
()
()
cos
2
()
() ()
θ
1
+
=
⇔+
1cot
2
() ()
θ
=
csc
2
θ
.
sin
2
θ
sin
2
θ
sin
2
θ
QED.
5. Prove the following,
tan(
θβ
+= −
)
tan
θ β
+
tan
tan(
θβ
−= +
)
tan
θ β
tan
1tantan
θ β
1tantan
θ β
a.
b.
cot(
θβ
+=
)
cot cot
β θ
−
1
cot(
θβ
−=
)
cot cot
β θ
+
1
cot
β θ
+
cot
cot
β θ
cot
c.
d.
.
5a.
1
( )
( )
tan
( )
+=
sin
θβ θ β θ β
=
sin cos
+
cos sin
=
cos cos
θβ
⋅
sin cos
θ β θ β
+
cos sin
cos
θ φ θ β θ β
cos cos
−
sin sin
1
cos cos
θ β θ β
−
sin sin
cos cos
θβ
sin cos
θβ θβ θ β
θβ θβ θ β
θβ θβ θ
θβ θβ
+
cos sin
sin
+
sin
cos cos
cos cos
cos
cos
tan
θ β
+
tan
=
=
=
cos cos
sin sin
1tant
an
β
1 tan tan
−
θ β
−
cos cos
cos cos
QED.
The trick to this proof is using the addition formulas for sine and cosine, and also finding
a clever expression of the number 1 to multiply by (recall we can always multiple by 1
without changing the expression), which allows us to do some simplifications. In this
example, the expression for 1, which we multiplied by, was
1
cos cos
θβ
= .
1
1
cos cos
θβ
Frank Luna
Page 3
www.gameinstitute.com
θ
−
−
+
θβ
+
−
736239423.002.png 736239423.003.png
Game Math Chapter 5 Solutions
5b. ( ) ( )
tan
θβ θ β
−= +− =
tan
( )
tan
θ β θ β
+− −
tan
( )
( )
=
tan
tan
.
1 tan tan
−
θ β
− +
1 tan tan
θ β
QED.
So the trick here is rewrite the subtraction as an addition with a negative angle so that we
can use the result from 5a. We also use the fact ( )
tan
−=− .
tan
The next two proofs are essentially analogous to the above two.
5c.
1
( )
( )
cot
( )
+=
cos
θβ θ β θ β
+
=
cos cos
−
sin sin
= ⋅
sin sin
θβ
cos cos
θ β θ β
−
sin sin
sin
θ β θ β θ β
+
sin cos
+
cos sin
1
sin cos
θ β θ β
+
cos sin
sin sin
θβ
cos cos
θβ θβ
−
sin sin
inin inin
sin cos
θβ θβ
θβ θβ β θ
θβ θβ
cot cot
θβ
−
1
=
=
cos sin
cot
+
cot
+
sin sin
sin sin
QED.
5d.
( )
( )
cot
( ) ( )
θβ θ β
−= +− =
cot
( )
cot cot
θ β
=
−+ − +
−− −
1
cot cot
θβ
−
1
cot
β θ
cot
cot
β θ
cot
=− ⋅
1
cot cot
θβ θβ
β θ β θ
+
1 cot cot
=
+
1
− +
cot
cot
−
cot
QED.
6. Derive formulae to rotate any point around the point ( x , y ), using the translation trick
mentioned in this lesson's material.
Recall the rotation equations to rotate a point ( )
x y counterclockwise about the origin:
,
′ = −
′ = +
cos
θ θ
θ θ
y
sin
yx
sin
y
cos
Frank Luna
Page 4
www.gameinstitute.com
β β
θβ
cot
xx
736239423.004.png
 
Game Math Chapter 5 Solutions
Here we wish to rotate a point ( )
ab , not about the origin, but about a pivot point
x y —see Figure 6a.
Figure 6a: We want to rotate a point (a, b) about a pivot point (x, y) into a new point (a’, b’).
x y is
the origin, and then translate the geometry so that its coordinates are specified relative to
this new coordinate system. This is easy enough, to transform points relative to ( )
,
x y
we offset their x coordinates by – x and offset their y coordinates by – y . Thus, ( )
,
ab
relative to ( )
x y is given by ( )
ab
aax
bby
Our geometry relative to this new coordinate system looks like Figure 6b.
Frank Luna
Page 5
www.gameinstitute.com
( )
,
Well, our rotation equations rotate about the origin of the coordinate system, so the first
thing we must do is to pick a new frame of reference (coordinate system) where ( )
,
:
=−
=−
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