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Trigonometry – Solutions
Revision Questions
1. Split the equilateral triangle into two right-angled triangles.
√ √ 2
a 3a 3a
Each has base and height , so the total area is .
2 2 4
2. From the graph of sin x, we expect two solutions in that range. If I draw a little triangle
with opposite side 1 and hypotenuse 2 then I recognise this as half an equilateral triangle
(so one solution is 30◦ , then using sin(180◦ − x) = sin x another is 150◦ ).
3. From the graph of tan x, we expect solutions to be 45◦ + 180◦ n for any whole number
n. In the given range, this is x = 45◦ or x = 225◦ .
4. If we write u = 45x then we’ve got tan(u) = 1 so u = 45◦ + 180◦ n just like in the
previous part. But careful, because if 45x = 45◦ + 180◦ n then x = 1◦ + 4◦ n, so the
solutions are the angles 1◦ , 5◦ , 9◦ , . . . , 357◦ , not just 1◦ and 5◦ .
5. Let’s draw a picture
√ 2
3
The area formulas give me the area as
1 √ 1 √ 1
× 1 × 3 × sin 90◦ or × 3 × 2 × sin 30◦ or × 1 × 2 × sin 60◦
2 2 2
√
all of which are equal to 3/2.
6. First I can expand the square to get cos2 x + 2 cos x sin x + sin2 x.
Now I can use cos2 x + sin2 x = 1 and I can write 2 cos x sin x as 2u.
So the expression is 2u + 1.
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7. This is a geometric series with first term a = 1 and common ratio r = − sin2 x, so the
sum to infinity is (1 + sin2 x)−1 which is (2 − cos2 x)−1 in terms of cos x.
The sum only converges if the common ratio r satisfies |r| < 1. This is not the case if
x = 90◦ .
8. I’ll use cos2 x = 1 − sin2 x and substitute this into the expression.
cos4 x + cos2 x = (1 − sin2 x)2 + (1 − sin2 x) = 2 − 3 sin2 x + sin4 x.
9. cos(450◦ − x) = cos(90◦ − x) using the fact that cos x is periodic with period 360◦ .
Then cos(90◦ − x) = sin x.
10. We can use cos(90◦ − x) = sin x and sin(90◦ − x) = cos x, and also sin(180◦ − x) = sin x
and cos(180◦ − x) = − cos x.
We have (sin x)(sin x) − (cos x)(− cos x) = sin2 x + cos2 x = 1.
√
11. Use the cosine rule: |AC|2 = 32 + 22 − 2 × 3 × 2 × cos 120◦ , so |AC| = 19.
√
× 3 × 2 × sin 120◦ = 3 2 3 .
Use the 21 ab sin γ formula for the area: 1
2
√
sin 120◦ sin ∠BCA 3 3
Use the sine rule: √ = so sin ∠BCA = √
19 3 2 19
12. The cosine rule is a2 = b2 + c2 − 2bc cos α.
Since −1 ⩽ cos α ⩽ 1, we have b2 − 2bc + c2 ⩽ b2 + c2 − 2bc cos α ⩽ b2 + 2bc + c2 .
So (b − c)2 ⩽ a2 ⩽ (b + c)2 . Since a > 0 and (b + c) > 0, it follows that a ⩽ b + c.
In fact, we can’t actually have a = b + c because that would be a triangle with a 180◦
angle. So a < b + c.
13. If we write down the cosine rule for each of the angles, we get
82 =132 +152 −2×13×15×cos α
132 =152 + 82 −2×15× 8×cos β
152 = 82 +132 −2× 8×13×cos γ
and these rearrange to
11 1 1
cos α = , cos β = , cos γ =
13 2 26
from which we see that β = 60◦ .
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MAT Questions
MAT 2014 Q1E
• Use cos2 x + sin2 x = 1 to write the function inside the brackets as 5 + 4 cos x − 4 cos2 x.
• Complete the square; this is −(2 cos x − 1)2 + 6. The maximum value of this quadratic
is 6, when cos x = 21 .
• Since −1 ⩽ cos x ⩽ 1, the bracket expression (2 cos x − 1) is between −3 and 1.
• So the value of 6 − (2 cos x − 1)2 could be as low as −3.
• The maximum value of this expression squared is 36 (from 62 , which is larger than
(−3)2 ).
• The answer is (b).
MAT 2015 Q1G
• Consider the graph of cos2 x. This has translational symmetries like cos2 (x + 180◦ ) =
cos2 x, and reflectional symmetry as cos2 x = cos2 (−x).
• So if cos2 x = cos2 y then we could have y = x, or y = x + 180◦ , or y = x + 360◦ , and
so on.
• Or we could have y = −x or y = 180◦ − x or y = 360◦ − x, and so on.
• These lines form a grid-like pattern, crossing the axes at multiples of 180◦ .
• The answer is (c).
MAT 2016 Q1D
• The substitution u = (cos x)n changes this equation to u + u2 = 0, with solutions u = 0
and u = −1.
• The first case corresponds to cos x = 0, which is true for x = 90◦ and for x = 270◦ and
no other values of x.
• The second case corresponds to cos x = −1 if n is odd, but has no solutions if n is even
(because then (cos x)n would be positive).
• So in that second case we gain the additional solution x = 180◦ , but only if n is odd.
• The answer is (d).
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MAT 2007 Q4
(i) The centre of the circle is at (1, 1), because the radius is 1, and that’s the distance from
the centre to each axis (measuring from the centre to each axis).
I drew a line from Q to the centre of the circle.
θ
x
P
I’ve called the angle that line makes with the positive-x direction ϕ. Because the tangent
is at right angles to the radius, we have ϕ = 90◦ − θ. A little bit of trigonometry, using
the fact that the radius of the circle is 1, gives the coordinates of Q as (1+cos ϕ, 1+sin ϕ).
Then I can convert this to the expression in the question by using cos(90◦ − θ) = sin θ
and sin(90◦ − θ) = cos θ.
The gradient of P QR is the change in y divided by the change in x. If I write |P R| for
the length of the hypotenuse of the right-angled triangle OP R, then the gradient is
−|P R| sin θ
.
|P R| cos θ
This simplifies to − tan θ.
The line has gradient − tan θ and goes through (1 + sin θ, 1 + cos θ). So the equation of
the line is
y = − tan θ (x − 1 − sin θ) + 1 + cos θ
The x-intercept of this line is at
1 + cos θ
x = 1 + sin θ +
tan θ
and that’s the x-coordinate of P . The y-coordinate of P is just zero.
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(ii) Draw two diagrams, one with θ as in the question, and one with 90◦ − θ instead.
B(90◦ − θ)
B(θ)
A(90◦ − θ)
A(θ)
The second diagram is just a reflection of the first diagram. The angles match because
the angles in a triangle add up to 180◦ . So the region marked A(θ) in the first diagram
has the same area as the region marked B(90◦ − θ) in the second diagram.
We can use this to calculate A(45◦ ). In this case B(45) = A(90◦ − 45◦ ) = A(45◦ ) so the
A and B regions have the same area. My strategy now is to draw some lines from the
centre of the circle to the sides of the axes, and to write the total area of the triangle
as a square plus three-quarters of a circle plus A(45◦ ) plus B(45◦ ).
I’ll need to know the side length of the triangle.
√ This is just the x-coordinate of P ,
◦
which we found above. For θ = 45 , this is 2 + 2. So
√ 2 !
1 (2 + 2) 3π √ 3π
A(45◦ ) = −1− =1+ 2−
2 2 4 8
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(iii) For this part I’ll draw a line joining the centre of the circle to P .
The area A(60◦ ) is made of two triangles, minus the sector of a circle. Using the fact
that the radius is 1 and the 60◦ -angle at P is split into equal parts (congruent triangles)
1
by the line to the centre of the circle, we see that the area of each triangle is .
2 tan 30◦
π
The angle at the centre of the circle is 120◦ , so the area of the sector is . This gives
3
the final answer in the question.
Extension
• We can adapt the strategy above to find the area as
1 (180◦ − θ)
A(θ) = −π .
tan(θ/2) 360◦
Alternatively, use the x-coordinate of P to get the equivalent expression
1 + cos θ (180◦ − θ)
A(θ) = sin θ + −π .
tan θ 360◦
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MAT 2012 Q4
(i) The gradient is the change in y divided by the change in x. Here, we’re moving from
(0, 2) to (x, x2 ) so the change in y is x2 − 2 and the change in x is x − 0. The gradient
is
x2 − 2
.
x
(ii) The tangent to the parabola at B is the same thing as the tangent to the circle at
B, because the circle and parabola are tangential to each other. The tangent to the
parabola has gradient 2x. We found the gradient of the radius of the circle at B in the
previous part. The tangent is at right angles to the radius for a circle, and if two lines
are at right angles then their gradients multiply to give −1. So to find the x-coordinate
of B we need to solve 2
x −2
(2x) = −1.
x
This rearranges to 2x2 = 3 and we want the positive root for B (the negative root gives
A). Remember that the y-coordinate is the square of the x-coordinate because B lies
on the parabola.
(iii) We could aim to find an angle at the centre of the circle.
θ θ/2
A B A B
There are two ways we might do this, and both of them involve angles. We might find
all the lengths of the sides of the triangle in that diagram, and apply the cosine rule
to find θ. Or we might split the triangle in two by dropping a perpendicular from the
centre down to the opposite side. If we do that, we can find two side lengths which
define cos(θ/2). If you used the second method, you probably didn’t call the angle θ/2.
Either way, we need the distance from the centre to B, the radius of the circle. We
know the coordinates of both points, so this is just a matter of applying Pythagoras’
theorem to find the radius as
2 √
v
u r !2
u 3 3 7
r= t + −2 = .
2 2 2
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(iv) Returning to the algebra in parts (i) and (ii), we see that we need there to be solutions
to 2
x −a
(2x) = −1
x
This rearranges to 2x2 = 2a − 1, which has two distinct solutions (for A and B) if and
only if a > 12 .
If a ⩽ 12 then there are no solutions for A and B and the circle must instead be resting
on the parabola at the origin (the vertex of the parabola).
(v) In the first case, we can adapt some of the algebra above to find the radius as
v !2
u r 2 r
u 2a − 1 2a − 1 1
r=t + −a = a− .
2 2 4
In instead the circle is resting on the vertex, then p
the radius is just r = a because the
distance between the centre and the origin is just (0 − 0)2 + (a − 0)2 .
Extension
1 5
• cos 2θ = 2 −1 = − . The two expressions are equivalent; one of them just involves
7 7
an angle that is twice as large as the other.
• The result is very similar to the question on the Algebra sheet. A circle is the points
that area a particular distance from a point, so the circle that “rests” on the parabola
identifies the points that are at the minimum distance from the centre of the circle.
With the centre of the circle on the y-axis, thisis similar to the calculation we did
before with points above or below the point 0, 21 .
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