给定 sin A = 12/37,求 cos A 和 tan A
三角学是直角三角形的角和边之间的关系。在直角三角形中,有3个角,其中一个角是直角(90°),另外两个角是锐角,有3条边。与直角相对的一侧称为斜边。根据它们之间的角度,这些边之间有 6 个比率,它们被称为三角比。
6个三角比是:
- 正弦 (sin)
- 余弦 (cos)
- 切线(棕褐色)
- 割线 (cosec)
- 正割(秒)
- 余切 (cot)
正弦(sin):
角的正弦由与角和斜边相反的边的长度之比定义。对于上述三角形,sin A = BC/AB
余弦(cos):
角的余弦由与角和斜边相邻的边的长度之比定义。对于上述三角形,cos A = AC/AB
切线(tan):
角的正切定义为与角相对的边与与角相邻的边的长度之比。对于上述三角形,tan A = BC/AC
余割(cosec):
角的余割由斜边的长度与角对边的比值定义。对于上述三角形, cosec A = AB/BC
割线(秒):
角的割线定义为斜边的长度与与角相邻的边和边的比值 对于上述三角形,sec A = AB/AC
余切(cot):
角的余切定义为与角相邻的边与对角的边的长度之比。对于上述三角形,cot A = AC/BC
给定 sin A = 12/37,求 cos A 和 tan A。
解决方案:
It is given that,
sinA = 12/37 ………………….( 1 )
We know that, cos2X = 1 – sin2X
⇒ cos2A = 1 – sin2A
⇒ cos2A = 1 – (12 / 37)2
⇒ cos2A = 1 – (144 / 1369)
⇒ cos2A = (1369 – 144) / 1369
⇒ cos2A = 1225 /1369
⇒ cosA = √1225 / √1369
⇒ cosA = 35 / 37 ………………….( 2 )
tanA = sinA/cosA
From (1) and (2),
⇒ tanA = (12/37) / (35/37)
⇒ tanA = 12/35
Therefore, the values of cosA and tanA are (35/37) and (12/35) respectively.
类似问题
问题 1:如果 sinA = 3/5,求 cosA 和 tanA
解决方案:
It is given that,
sinA = 3 / 5 ⇢ (1)
We know that, cos2X = 1 – sin2X
⇒ cos2A = 1 – sin2A
⇒ cos2A = 1 – (3/5)2
⇒ cos2A = 1 – (9 / 25)
⇒ cos2A = (25 – 9) / 25
⇒ cos2A = 16 / 25
⇒ cosA = √16 / √25
⇒ cosA = 4 / 5 ⇢ (2)
tanA = sinA / cosA
From (1) and (2),
⇒ tanA = (3/5) / (4/5)
⇒ tanA = 3 / 4
Therefore, the values of cosA and tanA are (4/5) and (3/4) respectively.
问题 2:如果 Sin A = 3/4,计算 cos A 和 tan A。
解决方案:
It is given that,
sinA = 3 / 4 ⇢ (1)
We know that, cos2X = 1 – sin2X
⇒ cos2A = 1 – sin2A
⇒ cos2A = 1 – (3/4)2
⇒ cos2A = 1 – (9/16)
⇒ cos2A = (16 – 9)/16
⇒ cos2A = 7/16
⇒ cosA = √7/√16
⇒ cosA = √7/4 ⇢ (2)
tanA = sinA/cosA
From (1) and (2),
⇒ tanA = (3/4)/(√7/4)
⇒ tanA = 3/√7
Therefore, the values of cosA and tanA are (√7/4) and (3/√7) respectively.
问题 3:如果 cosA = 12/13,求 sinA 和 tanA。
解决方案:
It is given that,
cosA = 12 / 13 ⇢ (1)
We know that, sin2X = 1 – cos2X
⇒ sin2A = 1 – cos2A
⇒ sin2A = 1 – (12/13)2
⇒ sin2A = 1 – (144 / 169)
⇒ sin2A = (169 – 144) / 16
⇒ sin2A = 25 / 169
⇒ sinA = √25 / √169
⇒ sinA = 5 / 13 ⇢ (2)
tanA = sinA / cosA
From (1) and (2),
⇒ tanA = (5/13) / (12/13)
⇒ tanA = 5/12
Therefore, the values of sinA and tanA are (5/13) and (5/3) respectively.