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The minimum value of 2 sin 6x+cos 6x

WebTrigonometry. Trigonometry (from Ancient Greek τρίγωνον (trígōnon) 'triangle', and μέτρον (métron) 'measure') is a branch of mathematics concerned with relationships between angles and ratios of lengths. The field emerged in the Hellenistic world during the 3rd century BC from applications of geometry to astronomical studies. WebCorrect option is A) By, Fundamental Properties of f(x) periodic function with Period T. f(x+T)=f(x) Let f(x)=cos 6x+sin 6x. assume from option smallest period of f(x)=π/2 ∴f(x+π/2)=f(x) ∴sin 6(x+π/2)+cos 6(x+π/2)=cos 6x+sin 6x. This shows that 2π is a fundamental period of f(x). Was this answer helpful? 0 0 Similar questions

The value of the integral ∫ (dx)/((1 + e^x)(sin^6x + cos^6x)), x ∈ (-π ...

WebProve f (x)sinx has a minimum. 1) If g(x) ≥ 0 then there are some points (when sin(x) = 0) for which g(x) = 0, and the infimum is attained in those points (e.g. x = 0 ). 2) If there is a point x0 for which g(x0) = −ϵ < 0 ... Are Exponential and Trigonometric Functions the Only Non-Trivial Solutions to F ′(x) = F (x+ a)? WebI have sin 2 x = 2 3 , and I'm supposed to express sin 6 x + cos 6 x as a b where a, b are co-prime positive integers. This is what I did: First, notice that ( sin x + cos x) 2 = sin 2 x + cos 2 x + sin 2 x = 1 + 2 3 = 5 3 . Now, from what was given we have sin x … pray i die lyrics https://anliste.com

Find Min. value of f(x) = sin 6 x + cos 6 x. - askIITians

WebMar 23, 2016 · Here is the expression: 2 ( sin 6 x + cos 6 x) − 3 ( sin 4 x + cos 4 x) + 1 The exercise is to evaluate it. In my text book the answer is 0 I tried to factor the expression, but it got me nowhere. algebra-precalculus trigonometry Share Cite Follow edited Mar 23, 2016 at 13:34 Jean-Claude Arbaut 22.7k 7 50 82 asked Mar 23, 2016 at 13:11 Planet_Earth WebJul 6, 2024 · The value of the integral π/2 ∫ −π/2 dx (1+ex)(sin6x+cos6x) ∫ − π / 2 π / 2 d x ( 1 + e x) ( s i n 6 x + c o s 6 x) is equal to (A) 2π (B) 0 (C) π (D) π/2 jee main 2024 Share It On 1 Answer +1 vote answered Jul 6, 2024 by GovindSaraswat (45.5k points) selected Jul 6, 2024 by Swetakeshri Best answer Correct option is (C) π WebSimplify -cos (6x)sin (6x) Mathway Calculus Examples Popular Problems Calculus Simplify -cos (6x)sin (6x) − cos(6x)sin(6x) - cos ( 6 x) sin ( 6 x) Nothing further can be done with this topic. Please check the expression entered or try another topic. −cos(6x)sin(6x) - cos ( … scollag road isle of man

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The minimum value of 2 sin 6x+cos 6x

Period of cos^6x + sin^6x is Maths Questions - Toppr

WebFree Minimum Calculator - find the Minimum of a data set step-by-step Webf (x) is minimum when (sin2 x cos2x) is maximum and maximum value of (sin2 x cos2x) is 1/2 when x =450 So, min. value of f (x) is 1/2 [ANS]. Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly.

The minimum value of 2 sin 6x+cos 6x

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WebSolve your math problems using our free math solver with step-by-step solutions. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. WebSep 20, 2024 · Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get …

WebSolve your math problems using our free math solver with step-by-step solutions. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. WebFeb 1, 2024 · sin^6x+cos^6x= =(sin^2x+cos^2x)^3–3*(sin^2x)*(cos^2x)*(sin^2x+cos^2x) = (1)^3–3*(sin^2x)*(cos^2x)*(1) = 1- 3*(sin^2x)*(cos^2x) {For the maximum value, The second term,[3*(sin^2x)*(cos^2x)] must be minimum, which is only when x is 0. Therefore the product of sine and cosine term will be equal to 0. } =1–0. Hence, the maximum value is 1.

WebIf 4 − sin 2 2 x − 4 sin 2 x sin 2 2 x + 4 sin 4 x − 4 sin 2 x cos 2 x = 9 1 and 0 &lt; x &lt; π, then the value of x is This question has multiple correct options Hard WebFeb 2, 2024 · Now use this formula: s = cos ( t) + cos ( 2 t) + cos ( 3 t) + ⋅ ⋅ ⋅ + cos ( n t) = sin n 2 t cos n + 1 2 t sin ( t 2) So you have to find minimum of this function: y = sin ( 10 x) cos ( 11 x) sin ( x) By plotting in Wolfram minimum of y is about -2.8. Share.

WebTo find the local maximum and minimum values of the function, set the derivative equal to 0 0 and solve. −2sin(x) = 0 - 2 sin ( x) = 0 Divide each term in −2sin(x) = 0 - 2 sin ( x) = 0 by −2 - 2 and simplify. Tap for more steps... sin(x) = 0 sin ( x) = 0 Take the inverse sine of both sides of the equation to extract x x from inside the sine.

scollard energy receivershipWebThe minimum value of [3^ (sin^6x)] + [3^ (cos^6x)] is: (Note: It is 3 raised to the power of [sin^6x] ) A) [ (2.3)^ (1/8)] B) [ (3.2)^ (1/8)] C) [ (2.3)^ (7/8)] D) [ (3.2)^ (1/8)] Expert Answer 100% (1 rating) Let [y =3^ (sin^6x) + 3^ (cos^6x)] In order to determine the minimum value of y, we have to find dy/dx first. pray i don\\u0027t alter it further memeWebHard Solution Verified by Toppr Correct option is A) We have sin 6x+cos 6x=(sin 2x) 3+(cos 2x) 3 =(sin 2x+cos 2x) 3−3sin 2xcos 2x(sin 2x+cos 2x) =1−3sin 2xcos 2x=1− 43.4sin 2xcos 2x=1− 43(sin2x) 2 ⇒ Maximum value of sin 6x+cos 6x is 1− 43×0=1 and minimum values is 1− 43×1= 41 Was this answer helpful? 0 0 Similar questions scollans drumshanboWebWhat is the minimum value of sin^6 (x) + cos^6 (x)? Ad by OnlineShoppingTools.com Prime is now $14.99 A Month, But Few Know This Free Savings Hack. Did you notice that your Amazon costs went up? Experts reveal what to do about it. Learn More All related (37) Sort Recommended Ved Prakash Sharma scollard drive peterboroughWebDetailed step by step solution for identity cos^2(6x)+sin^2(6x) pray i don\u0027t alter the deal furtherWebFeb 18, 2024 · Explanation: Use trig identity: cosa − cosb = − 2sin( a + b 2)sin( a − b 2) In this case: a +b 2 = 4x. a −b 2 = 2x. cos(2x) − cos(6x) = 2sin(4x).sin(2x) = 0. a. sin2x = 0 --> Unit circle gives --> 2x = kπ. x = kπ 2. pray i don\\u0027t alter the deal furtherWebJan 10, 2024 · ∫ / 2 0 x x(x)cos5(x)dx using the substitution that you mentioned, then, since, when x goes from 0 to π 2, sin(x) goes from 0 to 1, you get ∫1 0arcsin(t)(t6 − 2t8 + t10)dt = 4( − 5156 + 3465π) 2401245. So, (1) = 4 ( − 5156 + 3465π) 2401245. pray i don\u0027t alter it further meme