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prove\:\tan^2(x)-\sin^2(x)=\tan^2(x)\sin^2(x) \frac{d}{dx}(\frac{3x+9}{2-x}) (\sin^2(\theta))' \sin(120) \lim _{x\to 0}(x\ln (x)) \int e^x\cos (x)dx \int_{0}^{\pi}\sin(x)dx

In computes sine, cosine, tangent, and their reciprocal. Calculates arcsin, arccos, and arctan values inside inverse trigonometric equations.

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Use the form asin(bx−c)+ d a sin (b x - c) + d to find the variables used to find the amplitude, period, phase shift, and vertical shift. Find the amplitude |a| | a |. Find the period of sin(6x) sin …

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Take the inverse sine of both sides of the equation to extract x x from inside the sine. 6x = arcsin(0) 6 x = arcsin (0) Simplify the right side. Tap for more steps... Divide each term in 6x = …

Use the Double Angle identity: sin (2 x) = 2 sin (x) cos (x) = − cos (u) + 2 sin (u) cos (u)

$\sin 6x = 6\cos^5 x\sin x - 20\cos^3 x\sin^3 x +6\cos x\sin^5 x$ As you can see, this gets us both the identities for $\cos 6x$ and $\sin 6x$ simultaneously, but it does require you to know a bit …