F.4 附加數學 - 二倍角公式

Q: 若 sinA ＝ 1/3 且 A 是銳角, 試求 (cosA+cos 2A )/(1+sinA+sin 2A ) 的值
Sol:
sinA ＝ 1/3
sin2A ＝ 1/9
1 － cos2A ＝ 1/9
cos2A ＝ 8/9
cosA ＝ ± 2√2/3 ( cosA ＝ - 2√2/3 rejected )
( cosA ＋ cos 2A ) / ( 1 ＋ sinA ＋ sin 2A )
＝ ( cosA ＋ 2cos2A － 1 ) / ( 1 ＋ sinA ＋ 2sinAcosA )
＝ ( 2√2/3 ＋ 2 * 8/9 － 1 ] / ( 1 ＋ 1/3 ＋ 2 * 1/3 * 2√2/3 )
＝ ( 6√2/9 ＋ 7/9 ) / ( 12/9 ＋ 4√2/9 )
＝ ( 6√2 ＋ 7 ) / ( 12 ＋ 4√2 )
＝ ( 6√2 ＋ 7 ) ( 12 － 4√2 ) / [( 12 ＋ 4√2 ) ( 12 － 4√2 )]
＝ ( 72√2 ＋ 84 － 48 － 28√2 ) / ( 144 － 32 )
＝ ( 36 ＋ 44√2 ) / 112
＝ ( 9 ＋ 11√2 ) ∕ 28
Ans: ( 9 ＋ 11√2 ) ∕ 28
Send me a letter if any steps you don’t understand.

sinA=1/3
cos²A=1-(1/9)
cos²A=8/9
cosA=2(sqrt2)/3
(cosA + cos2A) / (1 + sinA + sin2A )
=(cosA+cos²A-sin²A)/(1+sinA+2sinAcosA)
=[2(sqrt2)/3+(8/9)-(1/9)]/[1+(1/3)+2(1/3)(2sqrt2)/3]
=[(6sqrt2+7)/9]/[(12+4sqrt2)/9]
=(6sqrt2+7)/4(3+sqrt2)
=(6sqrt2+7)(3-sqrt2)/4(3^2-2)
=(11sqrt2+9)/28

sinA= 1/3
sinA=k/3k
y=k, r=3k
cosA =x/r=k(10)^0.5/3k=(10)^0.5 / 3

(cosA + cos2A) / (1 + sinA + sin2A )
=(cosA+cos^2A-sin^2A)/(1+sinA+2sinA cosA)
=[(10)^0.5+3]/[4+(10)^0.5]