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

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{{iTest box|year=2006|num-b=16|num-a=18|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Problems/Problem U2|U2]] '''•''' [[2006 iTest Problems/Problem U3|U3]] '''•''' [[2006 iTest Problems/Problem U4|U4]] '''•''' [[2006 iTest Problems/Problem U5|U5]] '''•''' [[2006 iTest Problems/Problem U6|U6]] '''•''' [[2006 iTest Problems/Problem U7|U7]] '''•''' [[2006 iTest Problems/Problem U8|U8]] '''•''' [[2006 iTest Problems/Problem U9|U9]] '''•''' [[2006 iTest Problems/Problem U10|U10]]}}
  
 
[[Category:Introductory Trigonometry Problems]]
 
[[Category:Introductory Trigonometry Problems]]

Latest revision as of 17:27, 30 November 2018

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, Answer Key)
Preceded by:
Problem 16 		Followed by:
Problem 18
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