sin x cos x sin x
Lookingat the same unit circle you will find that cos(θ) and sin(θ) will give the X and Y coordinates respectively for the point on the unit circle that is at θ angle from the X axis. These functions where historically defined in terms of circles, in fact they come from the Sanskrit Jyā (sine) and koti-jyā (cosine), which where the names
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 )
Therelationships between the graphs (in rectangular coordinates) of sin(x), cos(x) and tan(x) and the coordinates of a point on a unit circle are explored using an applet. Definitions 1- Let x be a real number and P(x) a point on a unit circle such that the angle in standard position whose terminal side is segment OP is equal to x radians.(O is the origin of the system of axis used).
Tableof Integrals. Power of x. x n dx = x n+1 (n+1) -1 + C. (n -1) Proof. x -1 dx = ln|x| + C. Exponential / Logarithmic. e x dx = e x + C. Proof. b x dx = b x / ln (b) + C.
contoh id card panitia 17 agustus ke 78. 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 + cosx 〗/sin〖x − cosx 〗 Let f x = sin〖x + cosx 〗/sin〖x − cosx 〗 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
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
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sin x cos x sin x
