sin x cos x sin x

Answer(1 of 10): Method 1 Method 2 I hope it helps ! integralof sin^5 (3x)cos^3 (3x)dx arrow_forward I have to solve on [0,2π) sinx (tan (x/2))-sinx=0 tyty arrow_forward Integrate sin^2 (7x)-cos (x) arrow_forward Integrate sin 4x cos 5x + cos 4x sin 5x arrow_forward Find ∫cosh 2x sinh2 2x dx. arrow_forward cos x-sinxtanx= arrow_forward Recommended textbooks for you Mathematics For Machine Technology 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 0 So a question for a test I just had was integrate by substitution: Sin (x)e^Cos (x). I did something like this: Let u=Cos (x) du=-sin (x) dx. ∫sin (x)e^Cos (x) dx = ∫-e^u du. =∫-e^Cos (x) du. = -e ^cos (x) + c. 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. Hủy Hợp Đồng Vay Tiền Online. Trigonometry 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 . $\sin\sinx=\cos\pi/2-\sinx$, 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 $\cos\cosx>\cos\pi/2-\sinx=\sin\sinx$. 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 x cos x sin x