Concept: Trigonometric Ratios for Allied Angles: sin (-θ) = -sin θ. cos (-θ) = cos θ. sin (nπ + θ) = (-1) Arccosine is the inverse of the cosine function and thus it is one of the inverse trigonometric functions. Arccosine is pronounced as "arc cosine". Arccosine of x can also be written as "acosx" (or) "cos -1 x" or "arccos". If f and f -1 are inverse functions of each other, then f (x) = y ⇒ x = f -1 (y). So y = cos x ⇒ x = cos-1(y). y = sin–1x [–1,1] –π π, 2 2 y = cos–1x [–1,1] [0,π] y = cosec–1x R– (–1,1) –π π, –{0} 2 2 y = sec–1x R– (–1,1) [0,π] – π 2 y = tan–1x R –π π, 2 2 y = cot–1x R (0,π) Notes: (i) The symbol sin–1x should not be confused with (sinx)–1. Infact sin–1x is an If you have gone through double-angle formula or triple-angle formula, you must have learned how to express trigonometric functions of \(2\theta\) and \(3\theta\) in terms of \(\theta\) only. Explanation: Let a = cos−1x. Then x = cosa and tana = sina cosa = ± ( √1 − x2 x) Answer: ±( √1 − x2 x) . Answer link. . Let a=cos^ (-1)x. Then x= cos a and tan a=sina/cos a=+- (sqrt (1-x^2)/x) Answer:+- (sqrt (1-x^2)/x) . In other words, the domain of the inverse function is the range of the original function, and vice versa, as summarized in Figure 6.3.1. Figure 6.3.1. For example, if f(x) = sin x, then we would write f − 1(x) = sin − 1x. Be aware that sin − 1x does not mean 1 sin x. The following examples illustrate the inverse trigonometric functions: Previous Year Questions. If tan−1xtan−1⁡x + tan−1ytan−1⁡y = 2π32π3 , then cot−1xcot−1⁡x + cot−1ycot−1⁡y is equal t ∫sin(1/x)dx = -cos(1/x) + x∫cos(1/x)/x^2 dx We can then apply integration by parts again to the integral on the right-hand side. This results in an infinite series, which can be simplified using the Maclaurin series expansion of cos(1/x). Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. For math, science, nutrition, history We have to prove that 3cos−1x = cos−1(4x3 −3x) First consider RHS, cos−1(4x3 −3x) Let us take x =cosθ, we know that cos(3θ) = 4cos3θ−3cosθ. cos−1(4x3 −3x) =cos−1(4cos3θ−3cosθ) = cos−1 (cos(3θ))= 3θ. ∵ θ = cos−1x, cos−1(4x3 −3x) = 3cos−1x. ∴ LH S =RH S. 0 ≤ 3cos−1x ≤π. 0 ≤ cos1 x ≤π/3. ∴ c2Q8.