
Fundamental Identities - Trigonometry - Socratic
"The fundamental trigonometric identities" are the basic identities: •The reciprocal identities •The pythagorean identities
How do you solve cos x = x? - Socratic
2018年2月20日 · Use Newton's method to find: x ~~ 0.73908513322 Looking at the graphs of y = x and y = cos x we see that there is exactly one real solution, actually somewhere in (1/2, 1): graph{(y-x)(y-cos x) = 0 [-5, 5, -2.5, 2.5]} Typically for such an equation with mixed polynomial and trigonometric terms, there is no algebraic solution. We can use Newton's method to get a sequence of increasingly better ...
Solve #cos(cos(cos(x)))=sin(sin(sin(x)))# - Socratic
2017年11月20日 · See below. As can be seen the graphics for cos(cos(cos(x))) (blue) and sin(sin(sin(x))) (red) intersect in two points for -pi le x le pi NOTE: As we can easily verify 0.540302 approx cos(1) le cos(cos(cos(x))) le cos(cos(1)) approx 0.857553 and -0.745624 approx -sin(sin(1)) le sin(sin(sin(x))) le sin(sin(1)) approx 0.745624 and for x = pi/2 we have cos(cos(cos(pi/2))) < sin(sin(sin(pi/2))) so ...
How do you prove cos (pi-x) =- cosx? - Socratic
2016年1月31日 · Use the cosine subtraction formula: #cos(alpha-beta)=cos(alpha)cos(beta)+sin(alpha)sin(beta)# When applied to #cos(pi-x)#, this gives
What is the derivative of #y = cos(cos(cos(x)))#? - Socratic
(dy)/(dx)=-sin(cos(cos(x)))sin(cosx)sin(x) In order to differentiate a function of a function, say y, =f(g(x)), where we have to find (dy)/(dx), we need to do (a) substitute u=g(x), which gives us y=f(u).
cos(x)cos(2x)cos(4x)cos(8x)cos(16x) ... cos(2^n x) - Socratic
2018年2月8日 · We want to simplify . #cos(x)cos(2x)cos(4x)cos(8x)...cos(2^nx)# The trick is to use the double-angle identity repeatedly
Is it true that cos^(-1) x = 1/cos x ? + Example - Socratic
2016年8月16日 · False due to a clash of conventions. If n > 1 is a positive integer, then: cos^n x = (cos x)^n This is a convenience of notation, to avoid having to use parentheses to distinguish, for example: (cos x)^2 and cos (x^2) By convention we can write: cos^2 x and cos x^2 respectively, without ambiguity. However, in the case of -1, we have a clash of notation. If f(x) is a function, then f^(-1)(x) is ...
What is difference between 2cosx and cos2x? | Socratic
2018年5月16日 · 2cos x - cos 2x = -2cos^2x + 2 cos x - 1 # They're very different and your example is not correct. 2 cos x is twice the cosine of angle x. It will be between -2 and 2. cos2x is an abbreviation for cos(2x). It is the cosine of the angle 2x, two times the angle x. The value of cos 2x will be between -1 and 1. Before we think about double angles, let's remember the formula for the cosine of the ...
How do you solve cos2x= cosx from 0 to 2pi? - Socratic
2016年4月11日 · 0, (2pi)/3, (4pi)/3, 2pi Use the identity: cos 2x = 2cos^2 x - 1. The given equation transforms to: 2cos^2 x - cos x - 1 = 0. Solve this quadratic equation for cos x. Since a + b + c = 0, use shortcut. There are 2real roots: cos x = 1 and cos x = c/a = - 1/2. a. cos x = 1 --> x = 0 or x = 2pi b. cos x = - 1/2 ---> x = +- (2pi)/3 The co-terminal to arc - (2pi)/3 - …
How do you find the integral of #cos(mx)*cos(nx)#? - Socratic
2016年3月26日 · Your question is: #intcos(mx)cos(nx)dx# To simplify this, use the cosine product-to-sum formula, namely: #cos(A)cos(B)=1/2[cos(A-B)+cos(A+B)]#