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Simplify the ratio displaystylefrac5−5sin2x5−5cos2x\\displaystyle \\frac{5 - 5\\sin^2 x}{5 - 5\\cos^2 x}displaystylefrac5−5sin2x5−5cos2x.
Simplify the expression 5−5sin2x5 - 5\\sin^2 x5−5sin2x using a Pythagorean identity.
Express sqrt5−5sin2x\\sqrt{5 - 5\\sin^2 x}sqrt5−5sin2x in terms of cosx\\cos xcosx, indicating any absolute values required.
In a right triangle, the hypotenuse has length sqrt5\\sqrt{5}sqrt5 and the side opposite angle xxx has length sqrt5,sinx\\sqrt{5}\\,\\sin xsqrt5,sinx. Show that the length of the adjacent side is sqrt5−5sin2x\\sqrt{5 - 5\\sin^2 x}sqrt5−5sin2x and then simplify it.
Simplify the expression 5−5cos2bigl(x+tfracpi4bigr)5 - 5\\cos^2\\bigl(x + \\tfrac{\\pi}{4}\\bigr)5−5cos2bigl(x+tfracpi4bigr).
Express 5−5sin2(2x)5 - 5\\sin^2(2x)5−5sin2(2x) in terms of cos(4x)\\cos(4x)cos(4x).
Solve the equation sqrt5−5sin2x=3\\sqrt{5 - 5\\sin^2 x} = 3sqrt5−5sin2x=3 for xxx in [0,2pi)[0,2\\pi)[0,2pi).
Solve 5−5sin2x=cosx5 - 5\\sin^2 x = \\cos x5−5sin2x=cosx for xxx in [0,2pi)[0,2\\pi)[0,2pi).
Simplify the square of the sum bigl(sinx+sqrt5−5sin2xbigr)2\\bigl(\\sin x + \\sqrt{5 - 5\\sin^2 x}\\bigr)^2bigl(sinx+sqrt5−5sin2xbigr)2.
Evaluate the indefinite integral displaystyleint(5−5sin2x),dx\\displaystyle \\int (5 - 5\\sin^2 x)\\,dxdisplaystyleint(5−5sin2x),dx.
Differentiate f(x)=sqrt5−5sin2xf(x) = \\sqrt{5 - 5\\sin^2 x}f(x)=sqrt5−5sin2x with respect to xxx.
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