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Verify the identity sinx cotx - cosx divided by tanx equals 0?
(sin(x)cot(x) - cos(x))/tan(x)(Multiply by tan(x)/tan(x))sin(x) - cos(x)tan(x)(tan(x) = sin(x)/cos(x))sinx - cos(x)(sin(x)/cos(x))(cos(x) cancels out)sin(x) - sin(x)0
What is dydx of sincostanx?
If y = sin(cos(tan(x))) Using the chain rule: (f(g(x)))' = f'(g(x)).g'(x) Then dy/dx = cos(cos(tan(x))).-sin(tan(x)).sec2(x) = -cos(cos(tan(x))).sin(tan(x)).sec2(x) Unfortunately I don't think this can be simplified much more. ( sec = 1/cos )
How do you express sin x plus cos x divided by cos x in terms of tan x?
(sin x + cos x) / cosx = sin x / cos x + cosx / cos x = tan x + 1
Square root of 3 sinx plus cosx equals 0?
sqrt(3sin(x)=cos(x)=0 // Square both sides3sin(x) + cos(x) = 0 // subtract cos(x) from both sides3sin(x) = -cos(x) // rearrangesin(x)/cos(x) = -1/3 //sin(x)/cos(x) = tan(x)tan(x) = -1/3x = tan^-1(-1/3) == -18,43484882 // tan^-1(inverse tan)
Sec - cos equals tansin?
Prove that tan(x)sin(x) = sec(x)-cos(x) tan(x)sin(x) = [sin(x) / cos (x)] sin(x) = sin2(x) / cos(x) = [1-cos2(x)] / cos(x) = 1/cos(x) - cos2(x)/ cos(x) = sec(x)-cos(x) Q.E.D
