y= cos x sin x + sin x cos x Let u = cos x sinx and v = sinx cosx β y = u + v β dy dx = du dx + dv dx Now, u = cosx sinx Taking log on both side, we have, logu = sinx log cosx Differentiate w. r. t. x on both side, 1 u du dx = sinx 1 cosx-sin x + log cosx cosx β du dx = u-tan x sin x + cosxlog cos x β du dx = cosx sinx-tan x sin x
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 Β± β1 - sin2x Using complement or cofunction identity, cosx = sinΟ/2 - x sinx + cosx = sinx + sinΟ/2 - x Thus, the expression for sin x + cos x in terms of sine is sin x + sin Ο/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 Ο/2 - x.Transcript Misc 2 Prove that: (sin 3π₯ + sin π₯) sin π₯ + (cos 3π₯ - cos π₯) cos π₯ = 0 Lets calculate (sin 3x + sin x) and (cos 3x - cos x) separately We know that sin x + sin y = sin ( (π₯ + π¦)/2) cos ( (π₯ β π¦)/2) Replacing x with 3x and y with x sin 3x + sin x = 2sin ( (3π₯ + π₯)/2) cos ( (3π₯ β π₯)/2 ο»ΏTrigonometry Examples Popular Problems Trigonometry Simplify sinx-cosxsinx+cosx Step 1Apply the distributive 2Multiply .Tap for more steps...Step to the power of .Step to the power of .Step the power rule to combine and .
Sine(sin): Sine of an angle is defined by the ratio of lengths of sides which is opposite to the angle and the hypotenuse. For the above triangle, the value of angle sine is given for both β A and β B, the definition for sine angle is the ratio of the perpendicular to its hypotenuse.
2 Answers Please see two possibilities below and another in a separate answer. Explanation Using Pythagorean Identity sin^2x+cos^2x=1, so cos^2x = 1-sin^2x cosx = +- sqrt 1-sin^2x sinx + cosx = sinx +- sqrt 1-sin^2x Using complement / cofunction identity cosx = sinpi/2-x sinx + cosx = sinx + sinpi/2-x I've learned another way to do this. Thanks Steve M. Explanation Suppose that sinx+cosx=Rsinx+alpha Then sinx+cosx=Rsinxcosalpha+Rcosxsinalpha =Rcosalphasinx+Rsinalphacosx The coefficients of sinx and of cosx must be equal so Rcosalpha = 1 Rsinalpha=1 Squaring and adding, we get R^2cos^2alpha+R^2sin^2alpha = 2 so R^2cos^2alpha+sin^2alpha = 2 R = sqrt2 And now cosalpha = 1/sqrt2 sinalpha = 1/sqrt2 so alpha = cos^-11/sqrt2 = pi/4 sinx+cosx = sqrt2sinx+pi/4 Impact of this question 208126 views around the world1β₯ sin(x)/x β₯ cos(x) Hang on, hang on. We are almost done. In the inequality, all of the terms represent functions. Therefore, we can say that f(x) = 1, g(x) = sin(x)/x, and h(x) = cos(x). We
$\sin\sinx=\cos\pi/2-\sinx$, 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 $\cos\cosx>\cos\pi/2-\sinx=\sin\sinx$.
Example17 Prove that sinβ‘γ5x β γ2sin 3x +γβ‘sinβ‘x γ/πππ β‘γ5x β πππ β‘x γ = tan x Taking L.H.S. sinβ‘γ5x + γsin x β γβ‘2sinβ‘3x γ/πππ β‘γ5x β πππ β‘x γ = γ(sinγβ‘γ5x + γsin x) β γβ‘γ2 sinγβ‘3x γ/πππ β‘γ5x β πππ β‘x γ Solving numerator and denominator separately sin 5x + sin
sinx)*cos(x) Natural Language; Math Input; Extended Keyboard Examples Upload Random. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance, music
Solution We know the values of trigonometric functions for specific angles. So, we have. sin 120Β° = sin (2 Γ 60Β°) β sin 120Β° = 2 sin 60Β° cos 60Β° (Because 2 sin a cos a = sin (2a)) β sin 120Β° = 2 Γ β3/2 Γ 1/2. β sin 120Β° = β3/2. The formula can also be conversely used to find the value of 2 sin a cos a using sin 2a.
