May be Trigo 2

Geometry Level 4

If sin ⁡ 4 x a + cos ⁡ 4 x b = 1 a + b {\frac{\sin^{4}x}{a}+\frac{\cos^{4}x}{b}=\frac{1}{a+b}} then

sin ⁡ 8 x a 3 + cos ⁡ 8 x b 3 = f ( a , b ) {\frac{\sin^{8}x}{a^3}+\frac{\cos^{8}x}{b^{3}}}=f(a,b)

find f ( 1 , 1 ) f(1,1)


The answer is 0.125.

Tanishq Varshney by Tanishq Varshney , 23, India

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2 solutions

Chew-Seong Cheong
Jul 15, 2015

When a = 1 a=1 and b = 1 b=1 , then:

⇒ sin ⁡ 4 x + cos ⁡ 4 x = 1 2 ( sin ⁡ 2 x ) 2 + ( cos ⁡ 2 x ) 2 = 1 2 ( 1 − cos ⁡ 2 x 2 ) 2 + ( 1 + cos ⁡ 2 x 2 ) 2 = 1 2 1 − 2 cos ⁡ 2 x + cos ⁡ 2 2 x + 1 + 2 cos ⁡ 2 x + cos ⁡ 2 2 x = 2 2 cos ⁡ 2 2 x = 0 cos ⁡ 2 x = 0 ⇒ x = π 4 \begin{aligned} \Rightarrow \sin^4{x} + \cos^4{x} & = \frac{1}{2} \\ (\sin^2{x})^2 + (\cos^2{x})^2 & = \frac{1}{2} \\ \left( \frac{1-\cos{2x}}{2} \right)^2 + \left( \frac{1+\cos{2x}}{2} \right)^2 & = \frac{1}{2} \\ 1 - 2\cos{2x} + \cos^2{2x} + 1 + 2\cos{2x} + \cos^2{2x} & = 2 \\ 2\cos^2{2x} & = 0 \\ \cos{2x} & = 0 \\ \Rightarrow x & = \frac{\pi}{4} \end{aligned}

Now, we have:

f ( 1 , 1 ) = sin ⁡ 8 x + cos ⁡ 8 x = ( 1 2 ) 8 + ( 1 2 ) 8 = 1 8 = 0.125 \begin{aligned} f(1,1) & = \sin^8{x} + \cos^8{x} \\ & = \left(\frac{1}{\sqrt{2}} \right)^8 + \left(\frac{1}{\sqrt{2}} \right)^8 \\ & = \frac{1}{8} = \boxed{0.125} \end{aligned}

In response to the Challenge Master...

{ sin ⁡ 4 x a + cos ⁡ 4 x b = 1 a + b . . . ( 1 ) sin ⁡ 4 x b + cos ⁡ 4 x a = 1 a + b . . . ( 2 ) \begin{cases} \dfrac{\sin^4{x}}{a} + \dfrac{\cos^4{x}}{b} = \dfrac{1}{a+b} &...(1) \\ \dfrac{\sin^4{x}}{b} + \dfrac{\cos^4{x}}{a} = \dfrac{1}{a+b} &...(2) \end{cases}

( 1 ) − ( 2 ) : ( 1 a − 1 b ) sin ⁡ 4 x + ( 1 b − 1 a ) sin ⁡ 4 x = 0 \begin{aligned} (1)-(2): \space \left(\dfrac{1}{a} - \dfrac{1}{b}\right) \sin^4{x} + \left(\dfrac{1}{b} - \dfrac{1}{a}\right) \sin^4{x} & = 0 \end{aligned}

⇒ ( 1 a − 1 b ) sin ⁡ 4 x = ( 1 a − 1 b ) cos ⁡ 4 x sin ⁡ x = cos ⁡ x ⇒ x = π 4 ⇒ ( 1 2 ) 4 ( 1 a + 1 b ) = 1 a + b ( a + b ) 2 = 4 a b a 2 + 2 a b + b 2 = 4 a b a 2 − 2 a b + b 2 = 0 ( a − b ) 2 = 0 a = b \begin{aligned} \Rightarrow \left(\frac{1}{a} - \frac{1}{b}\right) \sin^4{x} & = \left(\dfrac{1}{a} - \frac{1}{ b}\right) \cos^4{x} \\ \sin{x} & = \cos{x} \\ \Rightarrow x & = \frac{\pi}{4} \\ \Rightarrow \left(\frac{1}{\sqrt{2}}\right)^4 \left( \frac{1}{a}+ \frac{1}{b} \right) & = \frac{1}{a+b} \\ (a+b)^2 & = 4ab \\ a^2 + 2ab + b^2 & = 4ab \\ a^2 - 2ab + b^2 & = 0 \\ (a-b)^2 & = 0 \\ a &= b \end{aligned}

⇒ x = π 4 , a = b \Rightarrow \boxed{x = \frac{\pi}{4}, a = b}

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Moderator note:

Any thoughts on how to prove the generalization that Prakhar Bindal brought up?

This Is Shorter Than Mine . I Derived For Any General Pair (a,b) And got the results as 1/(a+b)^3 .

Prakhar Bindal - 5 years, 11 months ago

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Can you share how you derived that result? It's interesting that such a result exists, and I wonder if there are any other explanations for it.

Calvin Lin Staff - 5 years, 11 months ago

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Mine is a very simple approach .

I Simply Replaced (sin)^2 = 1-(cos)^2 in the original condition given

And After A Bit Manipulation

It Yielded Me A Biquadratic Equation In terms of cosx

And to my suprise it turned out to be a perfect square !

whose roots were

(cosx)^2 = b/a+b

and we can apply the pythagorean identity to get

(sinx)^2 = a/a+b

Substituting these values in the expression asked

and manipulating it yielded me 1/(a+b)^3 .

This Result Is True For Any real pair(a,b).

where a+b cant be zero otherwise the original condition given to us will not be valid

Prakhar Bindal - 5 years, 11 months ago

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@Prakhar Bindal – Right.

The reason for that is because if se set sin ⁡ 2 x = s , cos ⁡ 2 x = c \sin ^2 x = s , \cos ^2 x = c , we have s + c = 1 s + c = 1 like you mentioned. Now, by Titu's Lemma ,

s 2 a + c 2 b ≤ ( s + c ) 2 a + b = 1 a + b \frac{ s^2 } { a } + \frac{ c^2 } {b} \leq \frac{ ( s + c) ^2 } { a + b } = \frac{1}{a+b}

Thus, we get that s a = c b \frac{s}{a} = \frac{c}{b} as the equality condition for the equation to hold.

Calvin Lin Staff - 5 years, 11 months ago

@Tanishq Varshney This problem seems to be badly phrased to me. The x x is extremely dependent on the a , b a, b (and only holds true for particular cases), whereas your statement seems to suggest that we have a trigonometric identity instead.

Calvin Lin Staff - 5 years, 11 months ago
汶良 林
Jul 29, 2015

a = b = 1 a = b = 1

s i n 4 x + c o s 4 x = 1 / 2 sin^{4}x + cos^{4}x = 1/2

( s i n 2 x + c o s 2 x ) 2 − 2 s i n 2 x c o s 2 x = 1 / 2 (sin^{2}x + cos^{2}x)^{2} - 2sin^{2}xcos^{2}x = 1/2

s i n 2 x c o s 2 x = 1 / 4 sin^{2}xcos^{2}x = 1/4

f ( 1 , 1 ) = s i n 8 x + c o s 8 x f(1, 1) = sin^{8}x + cos^{8}x

= ( s i n 4 x + c o s 4 x ) 2 − 2 s i n 4 x c o s 4 x = (sin^{4}x + cos^{4}x)^{2} - 2sin^{4}xcos^{4}x

= 1 / 4 − 2 × ( 1 / 4 ) 2 = 0.125 = 1/4 - 2×(1/4)^{2} = \boxed{0.125}

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