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What is the cosine of negative pi over 3?
Cos(Pi/3) is 1/2 so Cos(-Pi/3) ould be flipped over the x-axis. The answer is still 1/2.
What is x equal y sq over cos x pi?
Can you please claify if you mean x=y^2/ pi*cos(x) , or x=y^2/cos(pi), since they are very different sums.
What does cosx-sinx equals?
sqrt(2)*cos(x + pi/4) [with x in radians], or sqrt(2)*cos(x + 90°) [with x in degrees]
What is the derivative of cos pi x?
By the 'Chain Rule' dy/dx = dy/du X du/dx Y = Cos (pi*x). Let pi*x = u Y = Cos(u) u = pi*x dy/du = -Sin(u) du/dx = pi Hence dy/dx = dy/du X du/dx => ( Chain Rule) dy/dx = -Sin(u) X pi Substitute u for pi*x Hence dy/dx = -Sin(pi*x) X pi Tidying up dy/dx = -piSin(pix) Done!!!!
If x is obtuse and sin x equals b how do you show sin 2x is less than b?
you need to explain it better but with what i got i if sin x equals b obviously sin 2x is double of b hence sin 2x is more than b. {Not obvious at all, actually. And the above is false. Sin(Pi/2) = 1 and Sin(Pi)=0. But clearly 2 is not greater than 0. Contradiction.} Obtuse means Pi/2 < x < Pi So, sin(x) = b means b>0, because sin(y) > 0 if 0<y<Pi sin(2x)=2sin(x)cos(x) cos(x) < 0 because cos(Pi/2)= 0 and the derivative is negative there. Hence, sin(2x) = 2 sin(x) cos(x) = 2*b*(-K), where K is a positive constant Since b>0, -2Kb < b
