HKDSE Mathematics Question Review (2025-04-19)

Obviously c < e

Note: if cos(θ) = cos(2θ),

then d=e, but obviously d>e

You may also consider cos(2θ) = 1,

then contradiction also occurs.

Solution for (b)(ii):

Suppose a=b:

By sin(2θ) = 2sin(θ)cos(θ) = 2sin(θ)sin(π/2-θ),

we have 2a/d = 2(b/e)(c/d)=2(a/e)(c/d),

which implies that c=e,

which contradict the fact that e cos(θ) = c

where 0 <cos(θ) < 1 and 0 < θ< π/2, i.e. c≠e

So, a≠b

Solution for (b)(i)

Suppose θ=β, where both are acute angles,

then angle MKL = π/2 + θ,

and angle NML = π/2 - 2θ

Consider triangle MKL, by sine law:

d/sin(π/2 + θ) = e/sin(π/2 - 2θ)

d/cos(θ) = e/cos(2θ)

= b/sin(β)cos(2θ) = a/sin(θ)cos(2θ)

Since also d/cos(θ) = a/sin(θ),

we have cos(θ) = cos(2θ)

The only solution is θ=0,

which contradicts the fact that π/2>θ>0

So β≠θ

Solution

Consider the origin O (0,0),

a point R (a, b) where a>0 and b>0,

another point S (-a, b) and RO = r= SO >0.

Then we have P(a, 0) and Q (-a, 0)

Now we have angle POR = θ = angle QOS

So, angle ROS = π - 2θ

For triangle ROS, by applying sine law:

RS / sin(π - 2θ) = RO/ sin(θ) = SO / sin(θ)

2b / sin(2θ) = r / sin(θ)

sin(2θ)/2b = sin(θ)/r

sin(2θ) = 2sin(θ) (b/r) = 2sin(θ)cos(θ)

Question:

(a) By using Cartesian plane, show that:

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

(b) Consider a right-angled triangle MNL:

(i) Show that if a=b, then θ≠β

(ii) Shoe that if θ=β, then a≠b