trig proof question?

Question

if x = tan@ + sin@, y = tan@ - sin@

prove x^4 + y^4 = 2xy(8 + xy)


I tried just sub in the equations but ended up with something weird like having to prove tan^2@ - sin^2@ = tan^2@sin^2@

can anyone show me the steps?

Answer 1

I agree, the result would follow if you could prove that
tan^2 x - sin^2 x = tan^2 x sin^2 x

RHS = sin^4 x / cos^2 x
= (1 - cos^2 x)^2 / cos^2 x
= (1/cos x - cos x)^2
= sec^2 x + cos^2 x - 2
= sec^2 x + 1 - sin^2 x - 2
= (sec^2 x - 1) - sin^2 x
= tan^2 x - sin^2 x

Note that sin^2 x + cos^2 x = 1
divide by cos^2 x
tan^2 x + 1 = sec^2 x
So sec^2 x - 1 = tan^2 x

Answer 2

The equation x^4+y^4=2xy(8+xy) is equivalent to (x^2-y^2)^2=16xy. I'll prove the latter one instead.

x = tan@ + sin@ --> xcos@ = sin@ (1 + cos@) [A]
y = tan@ - sin@ --> ycos@ = sin@ (1 - cos@) [B]

Multiplying [A] and [B]:
xy cos^2@ = sin^2@ (1 - cos^2@) = sin^2@ (sin^2@)
Thus,
xy = sin^4@ / cos^2@ [C]

Square [A] and [B]
x^2 cos^2@ = sin^2@ (1 + 2cos@ + cos^2@)
y^2 cos^2@ = sin^2@ (1 - 2cos@ + cos^2@)
and subtract the resulting equations,
(x^2 - y^2) cos^2@ = sin^2@ (4cos@)
then square
(x^2 - y^2)^2 cos^4@ = 16 sin^4@ cos^2@
(x^2 - y^2)^2 = 16 (sin^4@ / cos^2@) [D]

Combining [C] and [D],
(x^2-y^2)^2=16xy

Answer 3

x + y = 2tan(a)
x - y = 2sin(a)

sin(a) = (x - y) / 2 ...(1)
tan(a) = (x + y) / 2 ...(2)

cos(a) = sin(a) / tan(a)
= (x - y) / (x + y) ...(3)

xy = tan^2(a) - sin^2(a)
= [sin^2(a) - sin^2(a)cos^2(a)] / cos^2(a)
= sin^2(a)(1 - cos^2(a)) / cos^2(a)
= sin^4(a) / cos^2(a)

Substituting for sin(a) and cos(a) from (1) and (3):
xy = [(x - y)^4 / 16] / [(x - y)^2 / (x + y)^2]
= [(x - y)^4 / 16] * [(x + y)^2 / (x - y)^2]
= (x - y)^2(x + y)^2 / 16

16xy = (x^2 - y^2)^2
= x^4 + y^4 - 2x^2y^2
x^4 + y^4 = 16xy + 2x^2y^2
= 2xy(8 + xy).