A. 1-cos2A
B. 2sinA
C. 1-cos^2 (2A)
D. 2sinAcosA
E. 1/sec2A
2006-08-10 07:42:27 · 8 answers · asked by cocoabundle 1 in Science & Mathematics ➔ Mathematics
D. sin2A = 2sinAcosA
This comes from the more general identity sin(A + B) = sinAcosB + sinBcosA. Thus sin(A + A) = sinAcosA + sinAcosA = 2sinAcosA.
2006-08-10 07:47:51 · answer #1 · answered by pingu2me 1 · 0⤊ 0⤋
D.2sinAcosA
2006-08-10 14:46:07 · answer #2 · answered by raj 7 · 0⤊ 0⤋
c.1-cos^2 (2A).d cannot be the answer cuz sinAcosA=sin2A when you multiply it by 2 you get 2sin2A
2006-08-10 14:49:35 · answer #3 · answered by T-bag 3 · 0⤊ 0⤋
The correct answer is D.
E actually evaluates to cos 2A, and C is actually the same as sin² 2A. In case you were curious. :)
Good luck! :-)
2006-08-10 14:48:24 · answer #4 · answered by Jay H 5 · 0⤊ 0⤋
It's 2sin(A)cos(A) because
sin(A+B)=sin(A)cos(B)+sin(B)cos(A)
Let A=B and you're done.
Doug
2006-08-10 14:49:32 · answer #5 · answered by doug_donaghue 7 · 0⤊ 0⤋
D
2006-08-10 15:03:37 · answer #6 · answered by Amar Soni 7 · 0⤊ 0⤋
D
2006-08-10 14:48:08 · answer #7 · answered by delmet 2 · 0⤊ 0⤋
D
2006-08-10 14:46:23 · answer #8 · answered by AresIV 4 · 0⤊ 0⤋