We start with
sin(π cos θ)
cos(π sin θ)
=
cos(π cos θ)
sin(π sin θ)
we cross multiply to get rid of the fractions.
This results in
sin(π cos θ) sin(π sin θ) = cos(π sin θ) cos(π cos θ)
Now, if we can, we recognise this looks like sum angle identity for cosine if we
bring the sin × sin part to the other side of the equation,
this results in
cos(π sin θ) cos(π cos θ) − sin(π cos θ) sin(π sin θ) = 0
From the sum angle identity for cosine,
cos(A + B) = cos A cos B − sin A sin B
we identify A = π sin θ and B = π cos θ.
This allows us to rewrite the equation as
cos(π sin θ + π cos θ) = 0
The zeroes for the cosine function are known. These are when the argument is
some odd multiple of π/2.
Therefore,
π sin θ + π cos θ = π2 (2n + 1)
where n is a natural number including zero.
We may divide out the π from
√ both sides of the equation. Now, we note that
sin θ + cos θ can not exceed 2. This is because it has a maximum at π/4.
On the interval where θ runs from (0, π/2), this means that we can only
consider the value n = 0. Other values are not mathematically meaningful at
this point.
So now, we have the equation
sin θ + cos θ = 1/2
The problem asks us for cos(θ − π/4).
We recognise again that we make use the sum-angle identity for cosine to get
it in a form similar to the equation we have.
Indeed,
√
cos(θ − π/4) = cos θ cos(π/4) + sin θ sin(π/4) = 22 cos θ + sin θ
Substituting this into the equation,
sin θ + cos θ = 1/2
we find that
√2
2
cos(θ − π/4) = 1/2
Whereupon
cos(θ − π/4) =
^
¨
√
2/4