Funky Trigonometric Expression

The first intuition I got when I saw this question was to apply the product-to-sum formula to convert $\sin{75} \cos{75}$ into $\frac{1}{2}(\sin{(75+75)} + \sin {(75-75)}) = \frac{1}{2}(\sin{150}) = \frac{1}{4}$ .

It follows that $3 \sin^2 {75} \cos^2 {75} = \frac{3}{16}, 4 \sin^2 {75} \cos^2 {75} = \frac{4}{16}$ .

The next important intuition to have is $(x^4 + y^4)(x^2 + y^2) = x^6 + y^6 + x^4y^2 + y^4x^2$ . Hence, $(\sin^4 {75} + \cos^4 {75} )(\sin^2 {75} +\cos^2 {75} ) = \sin^6 {75} + \cos^6 {75} + \sin^4 {75} \cos^2 {75} + \cos^4 {75}\sin^2 {75} = \sin^6 {75} + \cos^6 {75} + (\sin^2{75} + \cos^2{75})(\sin^2{75} \cos^2{75})$ .

Why do we want to do the above? Because of the identity $sin^2{x} + cos^2 {x} = 1$ ! Then it follows from above that: $\sin^4 {75} + \cos^4 {75} = \sin^6 {75} + \cos^6 {75} + \sin^2{75} \cos^2{75} = \sin^6 {75} + \cos^6 {75} + \frac{1}{16}.$

Hence our expression becomes:

$\displaystyle \frac{\cos^4{75} + \sin^4{75} + \frac{3}{16}}{\cos^4{75} + \sin^4{75} - \frac{1}{16} + \frac{4}{16}} = \boxed{1}$

Let us start with the denominator:

$\cos^6{75^\circ} + \sin^6{75^o} + 4\sin^2{75^o}\cos^2{75^o}$

$= (\cos^2{75^o} + \sin^2{75^o}) (\cos^4{75^o} + \sin^4{75^o}) - \sin^2{75^o}\cos^4{75^o} - \sin^4{75^o}\cos^2{75^o} + 4\sin^2{75^o}\cos^2{75^o}$

$= (1) (\cos^4{75^o} + \sin^4{75^o}) - \sin^2{75^o}\cos^2{75^o} ( \sin^2{75^o} + \cos^2{75^o}) + 4\sin^2{75^o}\cos^2{75^o}$

$= \cos^4{75^o} + \sin^4{75^o} - \sin^2{75^o}\cos^2{75^o} (1) + 4\sin^2{75^o}\cos^2{75^o}$

$= \cos^4{75^o} + \sin^4{75^o} + 3\sin^2{75^o}\cos^2{75^o} = \text{numerator}$

Therefore,

$\dfrac {\cos^4{75^o} + \sin^4{75^o} + 3\sin^2{75^o}\cos^2{75^o} } {\cos^6{75^o} + \sin^6{75^o} + 4\sin^2{75^o}\cos^2{75^o}}$

$= \dfrac {\cos^4{75^o} + \sin^4{75^o} + 3\sin^2{75^o} \cos^2{75^o} } {\cos^4{75^o} + \sin^4{75^o} + 3 \sin^2{75^o} \cos^2{75^o}} = \boxed{1}$

That was simple the numerator and denominator simplifies to the same expression by basic algebra.... I mean there is no need of product to sum formula just take the numerator as 1+sin^2 (75) cos^2 (75) by $(a+b)^2$ Similarly for denominator use $(a+b)^3$ formula where a and b are sin^2 (75) and cos ^2 (75)...sorry I am working on phone so I found formatting complicated....

Let $a = \cos^2 75^{\circ}$ and $b = \sin^2 75^{\circ}$ . Then the expression is of the form $\frac{a^2 + b^2 + 3ab}{a^3 + b^3 + 4ab}.$ Applying the sum of cubes factorization to the first two terms in the denominator, we have $\frac{a^2 + b^2 + 3ab}{(a + b)(a^2 - ab + b^2) + 4ab}.$ But since $\sin^2 x + \cos^2 x= 1$ for any $x$ , the $(a + b)$ factor disappears, and we find $\frac{a^2 + b^2 + 3ab}{a^2 + b^2 - ab + 4ab} = 1.$

using calculator ^^

Using algebra identity Cos^2 75 = a Sin^2 75 = b.... (a+b)^2 = a^2 + b^2 + 2ab ....... Then, cos^4 75 + sin^4 75 + 2 cos^2 75 sin^2 75 + cos^2 75 sin^2 75 = (sin^2 75 + cos^2 75)^2 + cos^2 75 sin^2 75...... The denominator...... (a+b)^3 = a^3 + b^3 + 3 ab (a+b)...... Cos^6 75 + sin^6 75 + 3 cos^2 75 sin^2 75 (sin^2 75 + cos^2 75)....... Soo, the dominator would be (cos^2 75 + sin^2 75)^3 + cos^2 75 sin^2 75...... Then, (sin^2 75 + cos^2 75)= 1...... Soo, 1 + sin^2 75 cos^2 75/1 + sin^2 75 cos^2 75 = 1...

Assume cos(75) as "C" and sin(75) as "S" Simplifying the numerator or rewriting it gives [(C^2)^2] + [(S^2)^2] + 2(S^2)(C^2) + (S^2)(C^2) => as the first three terms in the above expression are in the form of a^2 + b^2 + 2ab where a=C^2 and b=S^2 We rewrite the numerator(N) as N = [(C^2 + S^2) ^2] + (C^2)(S^2)............................eq(1) Similarly, the denominator is simplified as [(C^2)^3 + [(S^2)^3] + 3(C^2)(S^2) [(C^2) + (S^2)] + (S^2)((C^2) NOTE: Here we've multiplied the third term with the sum of the squares of cos(75) and sin(75) whose value is "1". So the denominator(D) can be rewritten as D = [(C^2) + (S^2)] ^3 +(S^2)(C^2).........................eq(2) Comparing eq(1) & eq(2), we know that the sum of the squares of cos and sin of same angle is always" 1". So the numerator and denominator expresses as N/D = [1+(C^2)(S^2)]/[1+(C^2)(S^2)] Hence the value is "1" & the option B is the right answer.

let cos 75=b , sin 75 =a

The numerator would become - (a^4+b^4+3(a^2.b^2))

The denominator would become - (a^6+b^6+4(a^2.b^2)) Which can be written as -( (a^2)^3+(b^2)^3+4(a^2.b^2)) Using the formula (a^3+b^3)=(a+b)(a^2+b^2-ab) we get (a^2+b^2)(a^4+b^4-a^2.b^2)+4(a^2.b^2) Since sin^2(75) + cos^2(75) = 1 Denominator becomes:- (1)(a^4+b^4-a^2.b^2)+4(a^2.b^2) a^4+b^4-a^2.b^2+4(a^2.b^2)

(a^4+b^4+3(a^2.b^2))

Hence the numerator and the denominator become the same, which gives 1 as answer.

We can write everything in terms of one variable: $a = \cos^2 75º$ , since $\sin^2 75º$ is just $1 - a$ . The fraction then simplifies to:

$\displaystyle{\frac{a^2+(1-a)^2+3a(1-a)}{a^3+(1-a)^3+4a(1-a)}}$ $\displaystyle{= \frac{a^2+1-2a+a^2+3a-3a^2}{a^3+1-3a+3a^2-a^3+4a-4a^2}}$ $\displaystyle{= \frac{-a^2+a+1}{-a^2+a+1}}$ $\displaystyle{= \boxed{1}}.$