# Difference between revisions of "2006 iTest Problems/Problem 17"

## Problem

Let $\sin(2x) = \frac{1}{7}$. Find the numerical value of $\sin(x)\sin(x)\sin(x)\sin(x) + \cos(x)\cos(x)\cos(x)\cos(x)$. $\text{(A) }\frac{2305}{2401}\qquad \text{(B) }\frac{4610}{2401}\qquad \text{(C) }\frac{2400}{2401}\qquad \text{(D) }\frac{6915}{2401}\qquad \text{(E) }\frac{1}{2401}\qquad \text{(F) }0\qquad$ $\text{(G) }\frac{195}{196}\qquad \text{(H) }\frac{195}{98}\qquad \text{(I) }\frac{97}{98}\qquad \text{(J) }\frac{1}{49}\qquad \text{(K) }\frac{2}{49}\qquad \text{(L) }\frac{48}{49}\qquad$ $\text{(M) }\frac{96}{49}\qquad \text{(N) }\pi\qquad \text{(O) }\text{none of the above}\qquad \text{(P) }1\qquad \text{(Q) }2\qquad$

## Solution

Notice that adding and subtracting $2\sin^2(x)\cos^2(x)$ in $\sin^4(x) + \cos^4(x)$ results in a factorable expression: $\sin^4(x) + 2\sin^2(x)\cos^2(x) + \cos^4(x) - 2\sin^2(x)\cos^2(x)$.

The first three terms can be factored into $(\sin^2(x) + \cos^2(x))^2$, which simplifies to $1$. Also, since $\sin(2x) = 2\sin(x)\cos(x) = \frac{1}{7}$, we know that $2\sin^2(x)\cos^2(x) = \tfrac12 \cdot 4\sin^2(x)\cos^2(x) = \tfrac12 \cdot (2\sin(x)\cos(x))^2 = \tfrac{1}{98}$.

Thus, the expression equals $1 - \tfrac{1}{98} = \boxed{\textbf{(I) } \tfrac{97}{98}}$.

## See Also

 2006 iTest (Problems) Preceded by:Problem 16 Followed by:Problem 18 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • U1 • U2 • U3 • U4 • U5 • U6 • U7 • U8 • U9 • U10
Invalid username
Login to AoPS