1/2 cos (cos^-1 (1/2)) . let cos^-1 (1/2)=A :.cosA=1/2=cos (pi/3) :.A=pi/3 or cos^-1 (1/2)=pi/3 So , cos (cos^-1 (1/2)) = cos (pi/3) =1/2 [Ans]

2018年4月13日 · pi/2 Cosine is an even function, meaning cos (-x)=cosx. So, cos (-pi/2)=cos (pi/2)=0 Then, we really want cos^-1 (0) = x -> cos (x)=0 -> x=pi/2 as the domain of the inverse cosine is [-1, 1].

Refer to Explanation cos^-1 (1/2) = 60^@ Trigonometric Ratios of Special Angles:

2017年6月20日 · How do you find the values of θ given cos θ = 1?

2016年10月22日 · Let's start by determining the derivative of y = cos^-1 (x). y = cos^-1 (x) cosy = x Through implicit differentiation, you should have: -siny (dy/dx) = 1 dy/dx = -1/siny Since siny = sqrt (1 - cos^2y), we can rewrite as: dy/dx= -1/sqrt (1 - cos^2y) Since x = cosy, we can substitute: dy/dx= -1/sqrt (1 - x^2) Now, let's differentiate e^ (2x). The derivative of any function of the …

2018年3月19日 · To prove that cos4x −sin4x = 1 −2sin2x, we'll need the Pythagorean identity and a variation on the Pythagorean identity: ⇒ cos2x +sin2x = 1 ⇒ cos2x = 1 − sin2x I'll start with the left-hand side and manipulate it until it looks like the right-hand side using these two identities: LH S = cos4x − sin4x LH S = (cos2x)2 − (sin2x)2

2016年3月18日 · =5pi/6 least value cos^-1 (cos (7pi/6))=cos^-1 (cos (pi+pi/6))=cos^-1 (-cos (pi/6))=cos^-1 (cos (pi-pi/6))=cos^-1 (cos (5pi/6))=5pi/6

2018年4月7日 · Keeping in mind that the domain of the inverse cosine is [-1, 1], the only solution to the above equation is x=pi/4 Thus, we get sin (pi/4)=sqrt2/2

2017年3月25日 · Let arccos(1 2√2) = A then cosA = 1 2√2 = √2 4 = 0.35355 and using tables or scientific calculator A = 69.3∘

2016年1月13日 · 4 (1-2x)sin (1-2x^2 Use the chain rule multiple times. The first issue is the cosine function. The chain rule states that d/dx [cos (u)]=-sin (u)*u' d/dx [cos (1-2x ...