WebThis algorithm computes the greatest common divisor (notation: (a, b) = gcd(a, b)) of two positive integers a, b. It proceeds by replacing the pair a, b (say, with a ≤ b) by r, a, where r is the remainder of the division of b by a. This procedure, which preserves the gcd, is repeated until we arrive at r = 0. Example. Compute gcd(12, 44). Web2 Feb 1975 · Let f(x) = sin x cos x = 1 2 sin 2x. We know -1 ≤ sin2x ≤ 1 ⇒ - 1 2 ≤ 1 2 sin2x ≤ 1 2. Thus the greatest and least value of f(x) are 1 2 and - 1 2 respectively.
The greatest and least value of sin x cos x are - Byju
Web3 May 2024 · What is the greatest value of sin x from 45 to 90? Maths keeps one mentally active. values of x from 45 to 90, cos x < sin x. At x = 45, cos x = sin x. Greatest value of sin x = 1.0 is when x = 90 deg and the greatest value of cos x = 1.0 is when x = 0 deg. How to find min value of sin θ + cosec θ? WebCompute the required values: Given: sin x cos x. Multiplying and dividing by 2, ⇒ 1 2 2 sin x cos x. ⇒ sin 2 x 2. We know that, -1 ≤ sin 2 x ≤ 1. Dividing by 2,-1 2 ≤ sin 2 x 2 ≤ 1 2. Hence, the greatest and least values are 1 2 and -1 2 respectively. Option B is the correct answer. industrial foregoing tools nuke
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Web18 Mar 2005 · With [tex]f(x)=x^{2}-x\sin x-\cos x [/tex], try using the intermediate value theorem to show you have at least one root in [itex](0,+\infty)[/itex], then use the fact that this is even (as dexter pointed out) to prove at least two roots. Next consider what Rolle's theorem will say about the derivative of f(x) if you have more than two roots. Webiv) Order the quantities calculated in i, ii, and iii, from least to greatest. v) Compare p p´ 3 2 q with p p 1 6 q. Since p p x q is on r´ 3 2, 1 6 s and p p´ 3 2 q and p p 1 6 q are (opposite/same) in sign, the Intermediate Value Theorem gives that p p x q has a zero between ´ 3 2 and 1 6. 6) For f p x q “ sin x x, note that f p x q is ... Web31 Jul 2016 · 5sinX + 3cosX we have to find the maximum and minimum value of the given expression. Concept Used: Maximum value of a.sinθ + b.cosθ is √(a 2 + b 2) Minimum value of a.sinθ + b.cosθ is {- √(a 2 + b 2)} Calculation: Here the given expression is 5sinX +3cosX. Comparing with a.sinθ + b.cosθ get, a = 5 and b = 3. Maximum value is √(5 2 ... log home decorated for christmas inside