If sin^4 theta/a + cos^4 theta/b = 1/ a+b; prove sin^8theta/ a^3 + cos^8 theta/ b^3 =
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What have you tried so far?
not much :/
well in 1/a+b I've written 1 as sin^2 theta + cos^2 theta/ a+b
then I squared both sides and cubed but no success so far...
Just to make sure I am follwing the original question as it is not clear since you did not apply brackets appropriately to clear any potential confusion, is this what we have:
If \(\displaystyle sin^4\left(\dfrac{\theta}{a}\right) + cos^4\left(\dfrac{\theta}{b}\right)=\dfrac{1}{a+b}\),
prove \(\displaystyle sin^8\left(\dfrac{\theta}{a^3}\right) + cos^8\left(\dfrac{\theta}{b^3}\right)=\dfrac{1}{(a+b)^3}\)
I think you got it wrong...here it is...
( sin4[FONT=MathJax_Math]θ[/FONT]/ a) + ( cos4[FONT=MathJax_Math]θ[/FONT]/b ) = ( 1/ a+b )
Prove ( sin8 [FONT=MathJax_Math]θ[/FONT] / a3) + ( cos8 [FONT=MathJax_Math]θ[/FONT] / b3 ) = 1/ ( a+b )3
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Any progress?? What have you tried?Soroban, you're up!
I really don't have a handle on this, but I will jot down some musings.Any progress?? What have you tried?
We can normalize however we want, for instance, multiply both sides by (a*b):
b sin^4(theta) + a cos^4(theta) = ab/(a+b)
How can we deal with arbitrary a and b?
Neither a nor b nor the sum (a+b) may be zero.
We don't know the signs of a and b - but could try to solve first assuming both positive (?).
cos^2(theta) = 1 - sin^2(theta)
cos^4(theta) = 1 - 2 sin^2(theta) + sin^4(theta)
cos^8(theta) = 1 - 4 sin^2(theta) + 6 sin^4(theta) - 4 sin^6(theta) + sin^8(theta)
D
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So as I see the problem:
If
\(\displaystyle \displaystyle \ \frac{sin^4(\theta)}{a} \ + \ \frac{cos^4(\theta)}{b} \ = \ \frac{1}{a+b}\)
Prove that:
\(\displaystyle \displaystyle \ \frac{sin^8(\theta)}{a^3} \ + \ \frac{cos^8(\theta)}{b^3} \ = \ \frac{1}{(a+b)^3}\)
Above is at least true for a = b = 1 and Θ = 45°
I tried something different... I wrote (1/a+b) as (sin2theta + cos2theta/ a+b) and then squared it and cubed individually to get the terms... but haven't reached the answer...
yup thats what we need to prove... no use of values