Communications Engineering MSc - Preliminary Reading
2
Trigonometry Solutions
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1) With all of these problems, there are several ways to do them – but perhaps the simplest way
to understand is to use the formulas for the sin(A + B), sin(A - B), cos(A + B) and cos(A - B):
i) Simplify: cos(π – x)
Using the formula for cos(A - B), we get:
cos(π − x) = cos π cos x + sin π sin x
and then noting that cos(π) = -1 and sin(π) = 0, this simplifies to:
cos(π − x) = − cos x
ii) Simplify: sin(2π – x )
Using the formula for sin(A - B), we get:
sin(2π − x) = sin 2π cos x − cos 2π sin x
and then noting that cos(2π) = 1 and sin(2π) = 0, this simplifies to:
sin(2π − x) = − sin x
iii) Simplify: tan(π + x )
There are several ways to do this. One is to note that tan(x) = sin(x)/cos(x), and
therefore:
tan(π + x) =
=
sin(π + x) sin π cos x + cos π sin x
=
cos(π + x) cos π cos x − sin π sin x
− sin x sin x
=
= tan( x)
− cos x cos x
iv) Simplify: 1 – cos(2x) / sin(x).
There are several possible answers, but for this one, I’ll use the cos2(x)+sin2(x)=1 result
as well, to express the result in terms of sin(x) only – this would speed up any calculation of
this result done by computer, or in hardware, since only a single trigonometric function would
have to be evaluated.
1−
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cos(2 x)
cos 2 ( x) − sin 2 ( x)
= 1−
sin( x)
sin( x)
= 1−
1 − sin 2 ( x) − sin 2 ( x)
sin( x)
= 1−
1 − 2sin 2 ( x)
sin( x)
= 1−
1
+ 2sin( x)
sin( x)
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Communications Engineering MSc - Preliminary Reading
v) Simplify: 4sin(x)cos2(x) – sin(3x)
This one has a very simple answer, but you can get there just by considering that
sin(3x) = sin(2x + x), and using the same formulas as before:
4sin( x) cos 2 ( x) − sin(3 x) = 4sin( x) cos 2 ( x) − sin(2 x) cos( x) − sin( x) cos(2 x)
(
= 4sin( x) cos 2 ( x) − ( 2sin( x) cos( x) ) cos( x) − sin( x) cos 2 ( x) − sin 2 ( x)
= 4sin( x) cos 2 ( x ) − 2sin( x) cos 2 ( x) − sin( x) cos 2 ( x) + sin 3 ( x)
= sin( x) cos 2 ( x) + sin 3 ( x)
(
= sin( x) cos 2 ( x) + sin 2 ( x)
)
= sin( x)
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2) If cos(2π / x) = 0.5, find all possible values of x.
The important point is to find all of the values – sometimes the most obvious solution
is not the right one to use. Here, we have:
x=
2π
cos
and all possible values of cos-1(0.5) are 2nπ ±
( 0.5 )
−1
π
3
where n is any integer. This comes from the
first positive solution π/3, then noting that cosine is an even function, so –π/3 must be a
solution as well, and since the function is periodic with a period of 2π, we can add as many 2π
as we like to the angle, and we’ll still get the same answer.
Therefore, we can write:
2π
π
= 2nπ ±
3
x
x=
2π
2nπ ±
π
=
1
6
=
n ± 1/ 6 6n ± 1
3
The largest possible value of x is 6, the smallest is –6, and there are an infinite number of
solutions in-between.
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3) Solve for x: sin(2x) = cos(x).
First use the formula for sin(2x):
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)
Communications Engineering MSc - Preliminary Reading
2sin( x) cos( x) = cos( x)
cos( x) ( 2sin( x) − 1) = 0
Therefore, either cos(x) = 0, or sin(x) = 0.5. (Don’t just divide by cos(x) – you’ll lose the
solutions where cos(x) = 0, and they might be the ones you want!)
If cos(x) = 0, then x = nπ +π/2 where n is any integer.
If sin(x) = 0.5, then x = (2nπ + π /6) or (2nπ + 5π /6).
So, the full solution is: x = nπ +π/2, x = (2nπ + π /6) or (2nπ + 5π /6) where n is any integer.
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4) A triangle has sides of length 2, 3 and 4. What is the angle between the two shortest sides?
Noting that the largest angle must be the one between the two shortest sides, application of the
cosine formula gives:
42 = 22 + 32 − 2.2.3cos θ
cos θ =
4 + 9 − 16
12
θ = cos −1 ( −0.25 ) = 1.823 rad
which is 104.5 degrees.
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5) Prove the formula for sin(A+B). (Easy for those who know about complex numbers, a bit
harder for everyone else.)
Using complex numbers:
{
}
sin( A + B) = ℑ exp ( j ( A + B ) )
= ℑ{exp ( jA ) exp ( jB )}
= ℑ{( cos A + j sin A )( cos B + j sin B )}
= ℑ{cos A cos B + j sin A cos B + j sin B cos A − sin A sin B}
= sin A cos B + sin B cos A
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6) Show that the cosine of 45 degrees must be equal to the sine of 45 degrees by considering
the lengths of the sides of a right-angled triangle with two equal sides. Hence, using the
formula for sin(2x) and given that the sine of ninety degrees is one, determine the sine of 45
degrees.
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Communications Engineering MSc - Preliminary Reading
If the lengths of the two shorter sides are equal, then sin(45) = opposite / hypotenuse must
equal cos(45) = adjacent / hypotenuse, since the opposite and adjacent sides are the same
length.
Given that sin(90) = 1, we can derive that:
sin(90 degrees) = 2sin(45 degrees) cos(45 degrees)
1 = 2sin 2 (45 degrees)
sin 2 (45 degrees) =
sin(45 degrees) =
1
2
1
2
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7) Starting from the formulas for cos(A+B) and cos(A-B), prove that:
 x+ y
 x− y
cos( x) + cos( y ) = 2 cos 
 cos 

 2 
 2 
Let x = A + B and y = A - B, then A =
x+ y
x− y
and B =
, so that:
2
2
cos( x) + cos( y ) = cos ( A + B ) + cos ( A − B )
= cos A cos B − sin A sin B + cos A cos B + sin A sin B
= 2 cos A cos B
 x+ y
 x− y
= 2 cos 
 cos 

 2 
 2 
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8) Derive a formula for the difference between two sines in terms of the product of two (or
more) cosine or sine functions:
sin( x) − sin( y ) = ?
This can be done in a similar way. Again, let x = A + B and y = A - B, then A =
B=
x− y
, so that:
2
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x+ y
and
2
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Communications Engineering MSc - Preliminary Reading
sin( x) − sin( y ) = sin ( A + B ) − sin ( A − B )
= sin A cos B + cos A sin B − sin A cos B + cos A sin B
= 2 cos A sin B
 x+ y  x− y
= 2 cos 
 sin 

 2   2 
There are similar formula for the sum of two sines and the difference between two cosines. It’s
not worth learning them all, but it is useful to know they exist, and how to work them out if
you need them.
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