Adv Alegbra HW Solutions 12

Adv Alegbra HW Solutions 12 - sec = sin cos sin 1 cos = 1 ,...

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12 1.24 Use double induction to prove that ( m + 1 ) n > mn for all m , n 0. Solution. According to Exercise 1.23, there are three things to verify. (i) S ( 0 , 0 ) : ( 0 + 0 ) 0 = 1 · 0. (ii) S ( m , 0 ) S ( m + 1 , 0 ) :if ( m + 1 ) 0 > m , then ( m + 2 ) 0 >( m + 1 ) · 0 = 0? (iii) S ( m , n ) S ( m , n + 1 ) : does ( m + 1 ) n > mn imply ( m + 1 ) n + 1 > m ( n + 1 ) ? Yes, because ( m + 1 ) n + 1 = ( m + 1 )( m + 1 ) n >( m + 1 ) mn = m 2 n + mn > mn + m , for m 2 n mn and mn m . Notice that 2 n > n is the special case S ( 0 , n ) . 1.25 For every acute angle θ , i.e., 0 <θ< 90 , prove that sin θ + cot θ + sec θ 3 . Solution. That θ is an acute angle implies that the numbers sin θ , cot θ , and sec θ are all positive. The inequality of the means gives £ 1 3 ( sin θ + cot θ + sec θ) ± 3 sin θ cot θ sec θ. Now sin θ cot θ
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Unformatted text preview: sec = sin cos sin 1 cos = 1 , so that 1 3 ( sin + cot + sec ) 3 1 and 1 3 ( sin + cot + sec ) 1 . Therefore, sin + cot + sec 3. 1.26 Isoperimetric Inequality. (i) Let p be a positive number. If 1 is an equilateral triangle with perimeter p = 2 s , prove that area (1) = s 2 / 27. Solution. This is an elementary fact of high school geometry....
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