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Show that 2tan(th) / (1+tan^2(th)) = sin(2th), where th = theta

We have 2tan(th) / (1 + tan^2(th)) = sin(2th) We know that tan(A) = sin(A) / cos(A), and 1 + tan^2(A) = sec^2(A) Therefore => (2sin(th) / cos(th)) / sec^2(th) => 2sin(th)*cos^2(th) / cos(th) => 2sin...
IC
Answered by Ian C. • Maths tutor
4202 Views

Use logarithms to solve the equation 3^(2x+1) = 4^100

We have 3^(2x+1) = 4^100 => log(3^(2x+1)) = log(4^100) => (2x+1)log(3) = 100log(4)
IC
Answered by Ian C. • Maths tutor
7186 Views

Use the binomial series to find the expansion of 1/(2+5x)^3 in ascending powers of x up to x^3 (|x|<2/5)

We want to rearrange the expression to the form (1+y)^n so we can use the general result: (1+y)^n=1+ny+[n(n-1)/2]y^2+[n(n-1)(n-2)/3!]y^3+... 1/(2+5x)^3 = (2+5x)^-3 = [2(1+5x/2)]^-3 = (2^-3)(1+5x/2)^-3 using ...
SJ
Answered by Saskia J. • Maths tutor
14731 Views

differentiate y=(5x-2)^5

25(5x-2)^4
MK
Answered by Miss K. • Maths tutor
4862 Views

Find the gradient of the curve (x^3)-4(y^2)=12xy at the point P(-8,8)

First of all differentiate the equation of the curve implicitly, giving: 3x 2 -8y(dy/dx)=12y+12x(dy/dx) =&gt; (dy/dx)(12x+8y)=3x 2 -12y =&gt; dy/dx=(3x 2 -12y)/(12x+8y) As dy/dx is the gradient of the curve,...
FG
4744 Views