Trigonometry Solutions

Part of the Oxford MAT Livestream.

Revision Questions

1. Split the equilateral triangle into two right-angled triangles.

An equilateral triangle split in two with a vertical line. A right angle is marked where that line meets the middle of the base of the triangle

Each has base a2 and height 3a2, so the total area is 3a24.

2. From the graph of sinx, 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(180x)=sinx another is 150).

3. From the graph of tanx, we expect solutions to be 45+180n 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+180n just like in the previous part. But careful, because if 45x=45+180n then x=1+4n, so the solutions are the angles 1,5,9,,357, not just 1 and 5.

5. Let's draw a picture

A right-angled triangle with hypotenuse 2 and other sides 1 and root-3.

The area formulas give me the area as 
12×1×3×sin90or12×3×2×sin30or12×1×2×sin60
all of which are equal to 3/2.

6. First I can expand the square to get cos2x+2cosxsinx+sin2x.

Now I can use cos2x+sin2x=1 and I can write 2cosxsinx as 2u.

So the expression is 2u+1.

7. This is a geometric series with first term a=1 and common ratio r=sin2x, so the sum to infinity is (1+sin2x)1 which is (2cos2x)1 in terms of cosx.

The sum only converges if the common ratio r satisfies |r|<1. This is not the case if x=90.

8. I'll use cos2x=1sin2x and substitute this into the expression.

cos4x+cos2x=(1sin2x)2+(1sin2x)=23sin2x+sin4x.

9. cos(450x)=cos(90x) using the fact that cosx is periodic with period 360.

Then cos(90x)=sinx.

10. We can use cos(90x)=sinx and sin(90x)=cosx, and also sin(180x)=sinx and cos(180x)=cosx.

We have (sinx)(sinx)(cosx)(cosx)=sin2x+cos2x=1.

11. Use the cosine rule: |AC|2=32+222×3×2×cos120, so |AC|=19.

Use the 12absinγ formula for the area: 12×3×2×sin120=332.

Use the sine rule: sin12019=sinBCA3 so sinBCA=33219

12. The cosine rule is a2=b2+c22bccosα.

Since 1cosα1, we have b22bc+c2b2+c22bccosαb2+2bc+c2.

So (bc)2a2(b+c)2. Since a>0 and (b+c)>0, it follows that ab+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+1522×13×15×cosα132=152+822×15×8×cosβ152=82+1322×8×13×cosγ
and these rearrange to
cosα=1113,cosβ=12,cosγ=126
from which we see that β=60.

MAT Questions

MAT 2014 Q1E

  • Use cos2x+sin2x=1 to write the function inside the brackets as 5+4cosx4cos2x.
  • Complete the square; this is (2cosx1)2+6. The maximum value of this quadratic is 6, when cosx=12.
  • Since 1cosx1, the bracket expression (2cosx1) is between 3 and 1.
  • So the value of 6(2cosx1)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 cos2x. This has translational symmetries like cos2(x+180)=cos2x, and reflectional symmetry as cos2x=cos2(x).
  • So if cos2x=cos2y 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=180x or y=360x, 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=(cosx)n changes this equation to u+u2=0, with solutions u=0 and u=1.
  • The first case corresponds to cosx=0, which is true for x=90 and for x=270 and no other values of x.
  • The second case corresponds to cosx=1 if n is odd, but has no solutions if n is even  (because then (cosx)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).

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.

The diagram from the question, with the radius to Q marked in, which makes an angle of phi with the horizontal

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=90θ. A little bit of trigonometry, using the fact that the radius of the circle is 1, gives the coordinates of Q as (1+cosϕ,1+sinϕ). Then I can convert this to the expression in the question by using cos(90θ)=sinθ and sin(90θ)=cosθ.

The gradient of PQR is the change in y divided by the change in x. If I write |PR| for the length of the hypotenuse of the right-angled triangle OPR, then the gradient is |PR|sinθ|PR|cosθ. This simplifies to tanθ.

The line has gradient tanθ and goes through (1+sinθ,1+cosθ). So the equation of the line is
y=tanθ(x1sinθ)+1+cosθ
The x-intercept of this line is at 
x=1+sinθ+1+cosθtanθ
and that's the x-coordinate of P. The y-coordinate of P is just zero.

(ii) Draw two diagrams, one with θ as in the question, and one with 90θ instead.

Left: the diagram from the question with A and B marked. Right: another picture of a slightly different triangle, with A(90-theta) and B(90-theta) marked

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(90θ) in the second diagram.

We can use this to calculate A(45). In this case B(45)=A(9045)=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).

An isosceles right-angled triangle, with a circle inside which is tangent to all three sides. A square is marked in, made of radii of the circle to the short sides of the triangle, and parts of the short sides themselves

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
A(45)=12((2+2)2213π4)=1+23π8

(iii) For this part I'll draw a line joining the centre of the circle to P.

The diagram in the question, but the centre of the circle is joined with lines to the points of contact and also 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) by the line to the centre of the circle, we see that the area of each triangle is 12tan30. The angle at the centre of the circle is 120, so the area of the sector is π3. This gives the final answer in the question.


Extension

  • We can adapt the strategy above to find the area as A(θ)=1tan(θ/2)π(180θ)360.
    Alternatively, use the x-coordinate of P to get the equivalent expression
    A(θ)=sinθ+1+cosθtanθπ(180θ)360.

 

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 x22 and the change in x is x0. The gradient is
x22x.

(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
(x22x)(2x)=1.
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.

Left: the triangle ABC is marked, where C is the centre of the circle. The angle at C is theta. Right: the same, but a perpendicular has been dropped down the middle of that triangle. The half-angle at C is marked theta over 2

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
r=(32)2+(322)2=72.

(iv) Returning to the algebra in parts (i) and (ii), we see that we need there to be solutions to
(x2ax)(2x)=1
This rearranges to 2x2=2a1, which has two distinct solutions (for A and B) if and only if a>12.

If a12 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
r=(2a12)2+(2a12a)2=a14.

In instead the circle is resting on the vertex, then the radius is just r=a because the distance between the centre and the origin is just (00)2+(a0)2.

 

Extension

  • cos2θ=2(17)1=57. The two expressions are equivalent; one of them just involves 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, this is similar to the calculation we did before with points above or below the point (0,12).
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