Sin20° = Cos 70 = 0.34. Cos 20 = Sin 70° = 0.94. Option D is the potential choice as for pair of angles 20° and 70Sin x = Cos y and Cos x = Sin Y. All others pairs of angles have not equal values for sin x and cos y. Answer: Option D 20°; 70° pair of angles has congruent values for the sin x° and the cos y°. הגבולשל sin (x)/x. ערך. שיחה. קריאה. גרסאות קודמות. כאשר ערכה של ה זווית x (ב רדיאנים) הולך ומתקרב לאפס, ה יחס בין ה סינוס של לבין הולך ומתקרב ל- . בלשון מתמטית, אומרים שה גבול של המנה כאשר שואף לאפס Idont know how to type cos^2(x)- sin^2(x) into matlab. I think it is cos(x).^2 - sin(x).^2. Also I know the interval is typed out [-2*pi, 2*pi} but I dont understand what the question means by saying using 100 points in the domain. John D'Errico on 23 Feb 2017. Inthis problem, both x and sin(2x) are not difficult to integrate so it is hard to know which one to choose as dv. It is tempting to pick x for dv because of how extremely easy it is to find the integral of x. For this very reason, I initially chose x for dv. But it turns out that the integral we get is not easier than the integral we had to Sinusund Kosinusfunktion (auch Cosinusfunktion) sind elementare mathematische Funktionen.Vor Tangens und Kotangens, Sekans und Kosekans bilden sie die wichtigsten trigonometrischen Funktionen.Sinus und Kosinus werden unter anderem in der Geometrie für Dreiecksberechnungen in der ebenen und sphärischen Trigonometrie benötigt. Auch in der Analysis sind sie wichtig. Let\\( f(x)=\\frac{\\sin ^{-1}(1-\\{x\\}) \\cos ^{-1}(1-\\{x\\})}{\\sqrt{2\\{x\\}}(1-\\{x\\} cdot\\} \\) denotes fractional part function) \\( \\frac{1}{2} \\) 1 2 RSokSR. 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 world 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 $\cos\cosx>\cos\pi/2-\sinx=\sin\sinx$. Misc 17 - Chapter 12 Class 11 Limits and Derivatives Last updated at May 29, 2023 by Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class Transcript Misc 17 Find the derivative of the following functions it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers sin⁡〖x + cos⁡x 〗/sin⁡〖x − cos⁡x 〗 Let f x = sin⁡〖x + cos⁡x 〗/sin⁡〖x − cos⁡x 〗 Let u = sin x + cos x & v = sin x – cos x ∴ fx = 𝑢/𝑣 So, f’x = 𝑢/𝑣^′ Using quotient rule f’x = 𝑢^′ 𝑣 −〖 𝑣〗^′ 𝑢/𝑣^2 Finding u’ & v’ u = sin x + cos x u’ = sin x + cos x’ = sin x’ + cos x’ = cos x – sin x v = sin x – cos x v’= sin x – cos x’ = sin x’ – cos x’ = cos x – – sin x = cos x + sin x Derivative of sin x = cos x Derivative of cos x = – sin x Now, f’x = 𝑢/𝑣^′ = 𝑢^′ 𝑣 −〖 𝑣〗^′ 𝑢/𝑣^2 = cos⁡〖𝑥 −〖 sin〗⁡〖𝑥 sin⁡〖𝑥 −〖 cos〗⁡〖𝑥 − cos⁡〖𝑥 +〖 sin〗⁡〖𝑥 sin⁡〖𝑥 +〖 cos〗⁡〖𝑥〗 〗 〗 〗 〗 〗 〗 〗/〖sin⁡〖x −co𝑠 𝑥〗〗^2 = −sin⁡〖𝑥 −〖 cos〗⁡〖𝑥 sin⁡〖𝑥 −〖 cos〗⁡〖𝑥 − sin⁡〖𝑥 + cos⁡〖𝑥 sin⁡〖𝑥 +〖 cos〗⁡〖𝑥〗 〗 〗 〗 〗 〗 〗 〗/〖sin⁡〖x − co𝑠 𝑥〗〗^2 = 〖−sin⁡〖x − co𝑠 𝑥〗〗^2 − 〖sin⁡〖x + co𝑠 𝑥〗〗^2/〖sin⁡〖x − co𝑠 𝑥〗〗^2 Using a + b2 = a2 + b2 + 2ab a – b2 = a2 + b2 – 2ab = − [sin2⁡〖𝑥 +〖 cos2〗⁡〖𝑥 − 2 sin⁡〖𝑥 〖 cos〗⁡〖𝑥 + 𝑠𝑖𝑛2𝑥 + 𝑐𝑜𝑠2𝑥 + 2𝑠𝑖𝑛𝑥 cos⁡〖𝑥]〗 〗 〗 〗 〗/〖sin⁡〖x − co𝑠 𝑥〗〗^2 = − 2𝑠𝑖𝑛2𝑥 + 2𝑐𝑜𝑠2𝑥 − 0/〖sin⁡〖x − co𝑠 𝑥〗〗^2 = −2 𝒔𝒊𝒏𝟐𝒙 + 𝒄𝒐𝒔𝟐𝒙/〖sin⁡〖x − co𝑠 𝑥〗〗^2 = −2 𝟏/〖sin⁡〖x − co𝑠 𝑥〗〗^2 = −𝟐 /〖𝒔𝒊𝒏⁡〖𝐱 − 𝒄𝒐𝒔 𝒙〗〗^𝟐 Using sin 2 x + cos 2 x = 1

sin x cos x sin x