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
Thearea bounded by the parabola y2 =8x y 2 = 8 x and its latus rectum in sq unit is Answer. 4. The area bounded by the curve y=sinx y = sin. . x between x=0 x = 0 and x=2π x = 2 π given by Answer. 5. The area bounded by the line y−x y − x , x-axis and lines x=−1 x = − 1 to x=2 x = 2, is Answer. 6. $\sin\sinx=\cos\pi/2-\sinx$, 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 $\cos\cosx>\cos\pi/2-\sinx=\sin\sinx$.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°.
Trigonometrysin(x −y) = sinxcosy −cosxsiny Similar Problems from Web Search Is it valid to write sin(x +iy) = sin(x)cos(iy)+ cos(x)sin(iy) Hint . Continuing what you did, and using the comment from Pedro Tamaroff, sin(x +iy) = sinxcoshy +icosxsinhy .Find: ∫ (sin x/ (sin3x + cos3x)) dx. integrals. indefinite integral. cbse. class-12. commented May 26 by jayasurya5764 (16 points) Cbse answer key = sarthaks.
sin x cos x sin x
