证明 1/(sin θ + cos θ) + 1/(sin θ – cos θ) = 2 sin θ/(1 – 2 cos 2 θ)
三角学是数学的一个分支,通常处理与其相关的三角形和角度。三角学有助于理解三角形的属性和三角形属性的实际应用。三角学由两个词组成,即三角和几何,它本身以简单的形式表达含义,即三角形的几何(其中三角指三角形)。
三角学有助于理解如何在给定三角形边长的情况下找到缺失的角度,或者它也可用于查找某些角度的值,如 0、30、45、60、90 等。
三角比
三角学中存在六个角度的函数或三角比。它们的名称和缩写是正弦 (sin)、余弦 (cos)、正切 (tan)、余切 (cot)、正割 (sec) 和余割 (csc)。这里要注意的一件重要事情是三角公式仅适用于直角三角形。让我们看看下图:
在这个直角三角形中,AC 边称为斜边。边 BC 被称为三角形的底边。 AB边称为三角形的高度。从上面的三角形中,我们可以列出上面的公式,
- sin∅ = AB/AC
- cos∅ = BC/AC
- tan∅ = AB/BC
- 婴儿床∅ = BC/AB
- cosec∅ = AC/AB
- 秒∅ = AC/BC
三角比的基本公式
- Tan ∅ = sin ∅/cos ∅
- sin ∅ = 1/cosec ∅
- cos ∅ = 1/秒 ∅
- 棕褐色 ∅ = 1/棉 ∅
- sin(90° – x) = cos x
- cos(90° – x) = sin x
- tan(90° – x) = 婴儿床 x
- 婴儿床 (90° – x) = 棕褐色 x
- 因为2 x+sin 2 x=1
- 1 + 婴儿床2 θ = cosec 2 θ
- 1 + tan 2 θ = 秒2 θ
证明 1/(sin θ + cos θ) + 1/(sin θ – cos θ) = 2 sin θ/(1 – 2 cos 2 θ)
证明:
Taking the LHS of the equation,
1/(sin θ + cos θ) + 1/(sin θ – cos θ)
= (sinθ – cos θ+sin θ + cos θ)/(sin2 θ – cos2 θ) [taking lcm of the equations]
= 2sinθ/(sin2 θ – cos2 θ)
in the RHS, the denominator is in cos terms. so, substituting sin2θ = 1 – cos2θ from the formulas,
= 2sinθ/(1- 2cos2 θ)
= RHS
类似问题
问题 1:证明 (1 – cos 2 θ)cosec 2 θ = 1
解决方案:
Taking the LHS of the equation,
(1-cos2 θ)cosec2θ
= (sin2θ)cosec2θ [as cos2x + sin2x = 1]
= cosec2θ /cosec2θ [as sin2θ = 1/cosec2θ]
= 1
= RHS
问题 2:证明 sec 2 θ + tan 2 θ = 2secθ – 1
解决方案:
Taking the LHS of the equation,
sec2θ + tan2θ
= 1/cos2θ + sin2θ/cos2θ [ as cos ∅ = 1/sec ∅ and Tan ∅ = sin ∅/cos ∅]
= (1+sin2θ)/cos2θ [taking lcm]
= (1+1-cos2θ )/cos2θ [as cos2x+sin2x = 1]
= (2-cos2θ )/cos2θ
= 2/cos2θ – cos2θ/cos2θ [splitting the denominator]
= 2sec2θ -1
= RHS
问题 3:证明 cotθ + cosecθ = (cosθ + 1)/sinθ
解决方案:
Taking the LHS of the equation,
cotθ + cosecθ
= cosθ/sinθ +1/sinθ [as sin ∅ = 1/cosec ∅ and cot∅ =cos∅ /sin∅ ]
= (cosθ+1)/sinθ [taking lcm]
= RHS
问题 4:证明 tanθ.sinθ + secθ.cos 2 θ = secθ
解决方案:
Taking the LHS of the equation,
tanθ.sinθ + secθ.cos2θ
= sinθ.sinθ/cosθ + cos2θ/cosθ [as as cos ∅ = 1/sec ∅ and Tan ∅ = sin ∅/cos ∅]
= (sin2θ + cos2θ)/cosθ [as cos2x+sin2x = 1]
= 1/cosθ
= secθ
= RHS