Transcript Misc 2 Prove that: (sin 3𝑥 + sin 𝑥) sin 𝑥 + (cos 3𝑥 - cos 𝑥) cos 𝑥 = 0 Lets calculate (sin 3x + sin x) and (cos 3x - cos x) separately We know that sin x + sin y = sin ( (𝑥 + 𝑦)/2) cos ( (𝑥 − 𝑦)/2) Replacing x with 3x and y with x sin 3x + sin x = 2sin ( (3𝑥 + 𝑥)/2) cos ( (3𝑥 − 𝑥)/2 sinx) cos(y) vs differentiate sin(x) cos(y) apply charcoal effect image of sin(x) cos(y) series sin(x) cos(y) plot 1/sin(x) cos(y) random vehicle curve; Have a question about using Wolfram|Alpha? Contact Pro Premium Expert Support » VIDEOANSWER:if you proved under course under minus X minus This I'm the minus X were coaching the course. I'm the X plus. Sign the X where we calm down, funder even identity here cause I'm the minus explain Recorded. Of course I off x because even something from the side off the Manus X because size Ah, what function Therefore we're gonna Manus on the X is in this even out identity we will Graphof y=sin (x) The graph of y=sin (x) is like a wave that forever oscillates between -1 and 1, in a shape that repeats itself every 2π units. Specifically, this means that the domain of sin (x) is all real numbers, and the range is [-1,1]. See how we find the graph of y=sin (x) using the unit-circle definition of sin (x). Thederivative of sin x with respect to x is cos x. It is represented as d/dx(sin x) = cos x (or) (sin x)' = cos x. i.e., the derivative of sine function of a variable with respect to the same variable is the cosine function of the same variable. i.e.,. d/dy (sin y) = cos y; d/dθ (sin θ) = cos θ; Derivative of Sin x Formula. The derivative of sin x is cos x. Theprocess to find the Taylor series expansion for {eq}sin (x) {/eq} will follow the same procedure used to find the Maclaurin series representation. First, find the derivatives of the given Proofof cos(x): from the derivative of sine. This can be derived just like sin(x) was derived or more easily from the result of sin(x). Given: sin(x) = cos(x); Chain Rule. Solve: cos(x) = sin(x + PI/2) cos(x) = sin(x + PI/2) = sin(u) * (x + PI/2) (Set u = x + PI/2) = cos(u) * 1 = cos(x + PI/2) = -sin(x) Q.E.D. Ifthe function f(x) = \\(\\frac{cos(sin\\,x)-cos\\,x}{x^4}\\) is continuous at each point in its domain and f(0) = \\(\\frac{1}{k},\\) then k is ______ . Derivativeof sin(x)*cos(x) - Answer | Math Problem Solver - Cymath \\"Get f(π/4)= 2cos2x. cos 3x-3 sin 2(cos 2(π/4)). cos(3(π/4))- 3 sin 3(π/4) . sin( π/4)= 2 cos 180. cos 135-3 sin 240.sin 45= 0 Beri Rating · 0.0 ( 0 ) nKnQ9em. Solution To convert sin x + cos x into sine expression we will be making use of trigonometric identities. Using pythagorean identity, sin2x + cos2x = 1 So, cos2x = 1 - sin2x By taking square root on both the sides, cosx + sinx = sinx ± √1 - sin2x Using complement or cofunction identity, cosx = sinπ/2 - x sinx + cosx = sinx + sinπ/2 - x Thus, the expression for sin x + cos x in terms of sine is sin x + sin π/2 - x. What is sin x + cos x in terms of sine? Summary The expression for sin x + cos x in terms of sine is sin x + sin π/2 - x. $\sin\sinx=\cos\pi/2-\sinx$, write $fx=\pi/2-\sinx-\cosx$, $f'x=-\cosx+\sinx$, we study $f$ in $[0,\pi/2]$, $f'x=0$ implies $x=\pi/4$, $f\pi/4>0$ $f0>0, f\pi/2>0$, implies that $f$ decreases from $0$ to $\pi/4$ and increases from $\pi/4$ to $\pi/2$, and $f>0$ on $[0,\pi/2]$. this implies that $\pi/2-\sinx>\cosx$, since $\cos$ decreases on $[0,\pi/2]$ we deduce that $\cos\cosx>\cos\pi/2-\sinx=\sin\sinx$.