sin 1 2 arctan x

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sin 1 2 arctan x We can think of arctan x 1 as a right triangle whose opposite is x and adjacent is 1 so the hypotenuse is sqrt 1 x 2 The sign of sine and of cosine are each ambiguous when we know the tangent

Simplify frac sin 4 x cos 4 x sin 2 x cos 2 x simplify frac sec x sin 2 x 1 sec x simplify sin 2 x cos 2 x sin 2 x simplify tan 4 x 2 tan 2 x 1 simplify tan 2 x cos 2 x cot 2 x sin 2 x Show More Step 1 Draw a triangle in the plane with vertices and the origin Then is the angle between the positive x axis and the ray beginning at the origin and passing through

sin 1 2 arctan x

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left sin x right 2 cdot left left cot x right 2 1 right Uh oh

The derivative of the arctan function is d dx arctan x 1 1 x The arctan function can be differentiated because its derivative exists at every point of its domain Looking at the graph of the single period of the function arctan x you Use transformations to sketch graphs of y frac 1 2 sin 1 x and y cos 1 left frac 1 2 x right Does frac 1 2 sin 1 x cos 1 left frac 1 2 x right Sketch a graph of y tan 1 x and label the scales on the axes

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Sin Tan and 1 are the heights to the line starting from the x axis while Cos 1 and Cot are lengths along the x axis starting from the origin Your original solution is correct as well It s just that x 45 circ doesn t satisfy sin 2 x frac14 If sin 2 x frac14 then sin x pm frac12 while sin 45 circ frac 1 sqrt 2

First let s call sin tan 1 x sin theta where the angle theta tan 1 x More specifically tan 1 x theta is the angle when tan theta x We know this from the definition of inverse functions For sin cos and tan the unit length radius forms the hypotenuse of the triangle that defines them The reciprocal identities arise as ratios of sides in the triangles where this unit line is no longer the hypotenuse

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sin 1 2 arctan x - Since frac x sqrt 1 x 2 0 we have sin arctan x frac x sqrt 1 x 2 in that case Similarly when x