Expert Answer to "If sin4θa+cos4θb=1a+b prove thatsin8θa3+cos8θb3=1(a+b)3"

Ellie Castro

Answered question

2022-04-21

\frac{\sin^{4} θ}{a} + \frac{\cos^{4} θ}{b} = \frac{1}{a + b}

prove that

\frac{\sin^{8} θ}{a^{3}} + \frac{\cos^{8} θ}{b^{3}} = \frac{1}{{(a + b)}^{3}}

haarplukxjf

Beginner2022-04-22Added 16 answers

Let

\frac{\sin^{4} θ}{a} = x

and

\frac{\cos^{4} θ}{b} = y

Cubing

(x + y)

{(x + y)}^{3} = x^{3} + y^{3} + 3 x y (x + y)

Substituting...

\frac{1}{{(a + b)}^{3}} = \frac{\sin^{12} θ}{a^{3}} + \frac{\cos^{12} θ}{b^{3}} + 3 \frac{\sin^{4} θ \cos^{4} θ}{a b} (\frac{\sin^{4} θ}{a} + \frac{\cos^{4} θ}{b})

Now equate this expression to

\frac{\sin^{8} θ}{a^{3}} + \frac{\cos^{8} θ}{b^{3}}

And LHS=RHS. Hence proved.

Eliza Flores

Beginner2022-04-23Added 16 answers

Let

\sin^{2} θ = k

Then our equation is.

\frac{k^{2}}{a} + \frac{{(1 - k^{2})}^{2}}{b} = \frac{1}{a + b}

I won't solve the whole equation but i will put in some steps. We will be getting something messy like:

k^{2} a b + k^{2} b^{2} + a^{2} + a^{2} k^{2} - 2 a^{2} k^{2} + a b + a b k^{2} - 2 a b k = a b

And the best part is this all simplifies to:

{(k \times (a + b) - a)}^{2} = 0

Hence

k = \frac{a}{a + b} = \sin^{2} θ

(1 - k) = \frac{b}{a + b} = \cos^{2} θ

Let us prove the general case.

\frac{\sin^{4 n} θ}{a^{2 n - 1}} + \frac{\cos^{4 n} θ}{b^{2 n - 1}} = \frac{1}{{(a + b)}^{2 n - 1}}

Substituting our values on LHS.

\frac{1}{a^{2 n - 1}} \times {(\frac{a}{a + b})}^{2 n} + \frac{1}{b^{2 n - 1}} \times {(\frac{b}{a + b})}^{2 n}

\frac{a}{{(a + b)}^{2 n}} + \frac{b}{{(a + b)}^{2 n}} = \frac{1}{{(a + b)}^{2 n - 1}} = R H S

Hence result is true.

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