simplify\:\frac{\sin^4(x)-\cos^4(x)}{\sin^2(x)-\cos^2(x)} simplify\:\frac{\sec(x)\sin^2(x)}{1+\sec(x)} \sin (x)+\sin (\frac{x}{2})=0,\:0\le \:x\le \:2\pi
sin^2(x)+cos^2(x) = 1 If sin(x) = 0.6 then (0.6)^2+cos^2(x) = 1 cos^2(x) = 1 - 0.36 = 0.64 cos(x) = +- 0.8
1) $\cos^2 x + \sin^2 x = 1$ So $2 \cos^2 x = 1$ So $\cos x = \sin x = \pm \sqrt{\frac 12}$ 2) $\sin x$ is the adjacent side of a right triangle.
Hence, the value of (sin 8x + 7sin 6x + 18 sin 4x + 12 sin 2x)/ (sin 7x+6 sin 5x+12 sin 3x) is 2 cos x. Stay tuned to BYJU’S – The Learning App and download the app to learn all Maths-concepts easily by exploring more videos.
Perhaps you should try it and see whether you get the series converging on the correct answer. for x = ½π you should get. cos x = 0. cos (x²) = −0.781211892. So, check a few terms of Maclaurin and see whether you are converging on those values. 2 comments.
Figure \(\PageIndex{3}\) shows the relationship between the graph of \(f(x)=\sin x\) and its derivative \(f′(x)=\cos x\). Notice that at the points where \(f(x
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what is cos x sin
