Trig. identities-problem??

Question

If xsin^3 theta+y cos^3 theta=sin theta. cos theta

& x sin theta=y cos theta

prove that x^2+y^2=1

Thanks :)))

Answer 1

Gee it took me like 25 mins to finally get this one out. You need to think of how to bring in the identity sin²Ɵ + cos²Ɵ = 1.

xsinƟ + ycos³Ɵ = sinƟcosƟ -------1

xsinƟ = ycosƟ ----------------------------2

So start with equation 1 and aim to eliminate y:

xsin³Ɵ + ycos³Ɵ = sinƟcosƟ

xsinƟsin²Ɵ + ycosƟcos²Ɵ = sinƟcosƟ

xsinƟsin²Ɵ + xsinƟcos²Ɵ = sinƟcosƟ   --------> as ycosƟ = xsinƟ

xsin²Ɵ + xcos²Ɵ = cosƟ   --------> after dividing through by sinƟ

x(sin²Ɵ + cos²Ɵ) = cosƟ

x = cosƟ --------> as sin²Ɵ + cos²Ɵ = 1.

Start with equation 1 again but now aim to eliminate x just as you did y above:

xsin³Ɵ + ycos³Ɵ = sinƟcosƟ

xsinƟsin²Ɵ + ycosƟcos²Ɵ = sinƟcosƟ

ycosƟsin²Ɵ + ycosƟcos²Ɵ = sinƟcosƟ -----> as xsinƟ = ycosƟ

ysin²Ɵ + ycos²Ɵ = sinƟ ------>after dividing through by cosƟ

y(sin²Ɵ + cos²Ɵ) = sinƟ

y = sinƟ  -------> as sin²Ɵ + cos²Ɵ = 1

So we found that:

y = sinƟ

x = cosƟ

Therefore we can say upon cubing both of these expressions that:

y³ = sin³Ɵ

x³ = cos³Ɵ

Now just put these four results back into equation 1:

xsin³Ɵ + ycos³Ɵ = sinƟcosƟ

x(y³) + y(x³) = xy

y² + x² = 1 -----> after dividing through by xy

x² + y² = 1

Hope this helped!

Answer 2

(1) xsin^3(θ) + ycos^3(θ) = sin(θ)cos(θ)

(2) xsin(θ) = ycos(θ)

From (2),

y = xsin(θ) / cos(θ)

Substitute into (1) :

xsin^3(θ) + [xsin(θ) / cos(θ)]cos^3(θ) = sin(θ)cos(θ)

Cancel cos(θ) from the second term :

xsin^3(θ) + xsin(θ)cos^2(θ) = sin(θ)cos(θ)

Divide through by sin(θ) :

xsin^2(θ) + xcos^2(θ) = cos(θ)

Take out the factor, x :

x[sin^2(θ) + cos^2(θ)] = cos(θ)

But sin^2(θ) + cos^2(θ) = 1 (identity)

Therefore, x = cos(θ)

Substitute into (2) to find that y = sin(θ)

Now, x^2 + y^2 = cos^2(θ) + sin^2(θ) = 1, as required.