三角学是数学的一个分支，涉及角度的研究，角度的测量以及测量单位。它还涉及给定角度的六个比率以及这些比率所满足的关系。从广义上讲，研究对象还包括构成三角形元素的角度。从逻辑上讲，是关于三角形性质的讨论；解决三角形，利用三角形的特性在高度和距离区域中的物理问题–均构成研究的一部分。它还提供了三角方程求解的方法。
三角恒等式
如果对角度的所有值均正确的方程，则包含该角度的三角比例的方程式称为“三角恒等式”。每当将三角函数包含在表达式或方程式中时，这些命令将很有用。六个基本的三角比例是正弦，余弦，正切，正割，割线和正切。使用直角三角形的边(例如相邻边，相对边和斜边)定义所有这些三角比。
三角恒等式的证明
对于任何锐角θ，证明
(i) tanθ = sinθ/cosθ (ii) cotθ = cosθ/sinθ (iii) tanθ . cotθ = 1
(iv) sin2θ + cos2θ = 1 (v) 1 + tan2θ = sec2θ (vi) 1 + cot2θ = cosec2θ
证明：
Consider a right-angled △ABC (fig. 1) in which ∠B = 90° and ∠A = 0°.
Let AB = x units, BC y units and AC = r units.
Then,
(i) tanθ = y/x = (y/r)/(x/r) [dividing num. and denom. by r]
∴ tanθ = sinθ/cosθ
(ii) cotθ = x/y = (x/r)/(y/r) [dividing num. and denom. by r]
∴ cotθ = cosθ/sinθ
(iii) tanθ . cotθ = (sinθ/cosθ) . (cosθ/sinθ)
tanθ . cotθ = 1
Then, by Pythagoras’ theorem, we have
x2 + y2 = r2.
Now,
(iv) sin2θ + cos2θ = (y/r)2 + (x/r)2 = ( y2/r2 + x2/r2)
= (x2 + y2)/r2 = r2/r2 = 1 [x2+ y2 = r2]
sin2θ + cos2θ = 1
(v) 1 + tan2θ = 1 + (y/x)2 = 1 + y2/x2 = (y2 + x2)/x2 = r2/x2 [x2 + y2 = r2]
(r/x)2 = sec2θ
∴ 1 + tan2θ = sec2θ.
(vi) 1 + cot2θ = 1 + (x/y)2 = 1 + x2/y2 = (x2 + y2)/y2 = r2/y2 [x2 + y2 = r2]
(r2/y2) = cosec2θ
∴ 1 + cot2θ = cosec2θ.
三角恒等式的应用
应用1：证明(1 – sin 2θ )sec 2θ = 1
证明：
We have:
LHS = (1 – sin2θ) sec2θ
= cos2θ . sec2θ
= cos2θ . (1/cos2θ)
=1
= RHS.
∴ LHS = RHS.
应用2：证明(1 + tan 2θ )cos 2θ = 1
证明：
We have:
LHS = (1 + tan2θ)cos2θ
= sec2θ . cos2θ
= (1/cos2θ) . cos2θ
= 1 = RHS.
∴ LHS=RHS.
应用3：证明(余割2θ – 1)tan²θ= 1
证明：
We have:
LHS = (cosec²θ – 1) tan2θ
= (1 + cot2θ – 1) tan2θ
= cot2θ . tan2θ
= (1/tan2θ) . tan2θ
= 1 = RHS.
∴ LHS=RHS.
应用4：证明(秒4θ -秒2θ)=(2黄褐色θ+黄褐色4θ)
证明：
We have:
LHS = (sec4θ – sec2θ)
= sec2θ(sec2θ – 1)
= (1 + tan2θ) (1 + tan2θ – 1)
= (1 + tan2θ) tan2θ
= (tan2θ + tan4θ)
= RHS
∴ LHS = RHS.
应用5：证明√(sec 2θ + cosec 2θ )=(tanθ+cotθ)
证明：
We have:
LHS = √(sec2θ + cosec2θ ) = √((1 + tan2θ) + (1 + cot2θ))
= √(tan2θ + cot2θ + 2)
= √(tan2θ + cot2θ + 2tanθ.cotθ ) (tanθ . cotθ = 1)
= √(tanθ + cotθ)2
= tanθ + cotθ = RHS
∴ LHS = RHS.