2006-10-21 21:07:50 · 10 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
1
2006-10-21 21:28:03 · answer #1 · answered by Anonymous · 0⤊ 1⤋
sinA + cosA
= 1. sinA + 1. cosA
= √2( 1/√2 sin A + 1/√2 cos A) (using √(1² + 1²) as the factor introduced)
= √2(sinA cos45° + cosAsin45°)
= √2sin(A + 45°)
Maximum value of sin(A + 45°) = 1 (at A = 45°) for 0 ≤ A ≤ 90°
Therefore maximum value of sinA + sinB
= maximum value of √2sin(A + 45°) for 0 ≤ A ≤ 90°
= √2 x 1
= √2
2006-10-21 21:28:51 · answer #2 · answered by Wal C 6 · 1⤊ 0⤋
y= sinA+cosA
dy/dA = cosA -sinA=0 at turning point
now,d^2y/dA^2 = -sinA-cosA which is -ve
therefore, dy/dA = cosA-sinA =0 is a maximum
>>>> cosA=sinA ......(1)
the only value to satisfy (1) for A in 0
giving cos45=sin45=1/((2)^(1/2))
therefore, max point for sinA+cosA = 1/((2)^(1/2))
+1/((2)^(1/2))
=(2)^(1/2)
(0
i hope that this helps
2006-10-21 22:17:37 · answer #3 · answered by Anonymous · 0⤊ 0⤋
to find maximum differentiate, put equal to zero and solve
diffrentiating sinA+cosA gives cosA-sinA, equating it to zero, gives sinA=cosA which is when A is 45 degrees
The value of sin 45=cos45=1/sqrt2
so Sin45+cos45= 2/sqrt2= sqrt2 = about 1.4
2006-10-21 23:05:12 · answer #4 · answered by grandpa 4 · 0⤊ 0⤋
d(sinA+cosA)/dA = cosA - sinA = 0 for max or min
sinA = cosA
A = 45°
(sqrt(2))/2 + (sqrt(2))/2 = sqrt(2)
2006-10-21 22:23:13 · answer #5 · answered by Helmut 7 · 0⤊ 0⤋
it happens in A=45 ( A= pi/4 )
Sin A + Cos A = Sqrt(2) / 2 + Sqrt(2) / 2 =Sqrt (2)
2006-10-21 21:14:26 · answer #6 · answered by Laughy kiLLer 1 · 0⤊ 0⤋
42!
2006-10-21 21:10:16 · answer #7 · answered by keoni_21 3 · 0⤊ 1⤋
sqrt2
differentiate and equate to zero.......... u get a = 45degrees.....
2006-10-21 21:10:03 · answer #8 · answered by !kumar! 2 · 1⤊ 0⤋
√2
Doug
2006-10-21 21:12:17 · answer #9 · answered by doug_donaghue 7 · 0⤊ 0⤋
$1.98
2006-10-21 21:14:40 · answer #10 · answered by Anonymous · 0⤊ 2⤋