Proving a trigonometric identity with ##\sin^4## s and ##\cos^4## s

[brotherbobby]

Homework Statement

If ##\dfrac{\sin^4 x}{a}+\dfrac{\cos^4 x}{b}=\dfrac{1}{a+b}##, then prove that

##\boxed{\boldsymbol{\frac{\sin^8 x}{a^3}+\frac{\cos^8 x}{b^3}=\frac{1}{(a+b)^3}}}##

Relevant Equations

The problem (identity) exists in the first chapter of the text, and, as such, all angles ##x## are acute and only the three basic identities involving squares can be assumed :
1. ##\sin^2 x+\cos^2 x=1##,
2. ##\sec^2 x = 1+\tan^2 x## and
3. ##\csc^2 x= 1+\cot^2 x##.

Problem statement : Let me copy and paste the problem as it appears in the text to the right.
Attempt :
Let me copy and paste my attempt. I couldn't go far, as you will see.

I couldn't progress from here. The powers of the ##\sin## and the ##\cos## are both what we want (##8##), but the denominators are squares and not the desired cubes. Also there's an extra term - what to do with it?

Request : A hint or help would be welcome.

 

[PeroK]

Request : A hint or help would be welcome.

Try multiplying ##\dfrac{\sin^4 x}{a} \times \dfrac{\cos^4 x}{b}##

 

[brotherbobby]

Try multiplying ##\dfrac{\sin^4 x}{a} \times \dfrac{\cos^4 x}{b}##

The original equation?

 

[PeroK]

The original equation?

To put it in simpler terms: what is:
$$\dfrac{\sin^4 x}{a} \times \dfrac{\cos^4 x}{b} = ?$$

 

[PeroK]

I have a different idea. Try finding ##\cos^2 x## i terms of ##a, b##.

 

[brotherbobby]

To put it in simpler terms: what is:
$$\dfrac{\sin^4 x}{a} \times \dfrac{\cos^4 x}{b} = ?$$

$$\dfrac{\sin^4 x}{a} \times \dfrac{\cos^4 x}{b} = \dfrac{\sin^4 x\cos^4 x}{ab}$$

 

[PeroK]

$$\dfrac{\sin^4 x}{a} \times \dfrac{\cos^4 x}{b} = \dfrac{\sin^4 x\cos^4 x}{ab}$$

I thought that would help, but as above there's a better idea.

 

[PeroK]

I thought that would help, but as above there's a better idea.

If you express ##\cos^2 x##, and hence also ##\sin^2 x##, in terms of ##a## and ##b##, the solution drops out immediately.

 

[PeroK]

More generally, we have:
$$\frac{\sin^{2n}x}{a^{n-1}} + \frac{\cos^{2n}x}{b^{n-1}} = \frac 1 {(a+b)^{n-1}}$$In the sense that, if that holds for ##n = 2##, then it holds for all ##n \ge 2##.

 

[anuttarasammyak]

Interesting exercise. I plot the graphs for a=1 b=2

where X = sin^2 x , Y = cos^2 x

PS a=1,b=-1/2

They seem to coincide at a point where cos^2 or sin^2 is minus.

 

[pasmith]

It is certainly worth simplifying things by setting [itex]\sin^2 x = Z[/itex] so that [itex]\cos^2 x= 1 - Z[/itex].

 

[anuttarasammyak]

Re:#10
The problem is restated as

Prove that all the three graphs
[tex]x+y=1...(1)[/tex]
[tex](1+p)x^2+(1+p^{-1})y^2=1...(2)[/tex]
[tex](1+p)^3x^4+(1+p^{-1})^3y^4=1...(3)[/tex]
intersect at a point (x,y) x>0 y>0. p=b/a > 0

You can get intersecting point of (1) and (2) by solving quadratic equation. Then check that this point is on (3) also.

 

[brotherbobby]

Thank you all for your comments and responses. I want to start by responding to @PeroK 's comment (#5) which I find leads to a solution. However, I need some time with @anuttarasammyak 's suggestions in comments 10 and 12 because I have, as of yet, to understand them.

I have a different idea. Try finding cos2⁡x i terms of a,b.

I work using ##\text{Mathtype}^{\circledR}##, hoping I am not violating something.