Trigonometric Functions (1 Viewer)

[dougal]

Would really appreciate the working to any of these questions... I have such trouble with trig.

1. Given d2y/dx2 = 9sin3x, find y if there is a stat pt at (pi/2, 1) [how do you get to y from its double derivative?]

2. Find lim as x->0 : 1 - cos(2x)/x(squared) [how do you work with x when it is small??]

3. Show that secxcosecx = sec(squared)x/tan(x) [this one i just couldn't figure out]

4. Find the general solutions of 2sin(cubed)x - sinx - 2sin(squared)x + 1 = 0 [how do you use general solutions?]

Thanks for any help... sorry for anything that doesnt quite fit! Not sure about some of these questions.

 

[hyparzero]

1. You integrate it twice

2. As x-approaches 0, cos(2x)/x² becomes extremely large ; ie. 1 / 0.0000001, so when x approaches zero, cos(2x)/x² approaches infinity. 1- infinity = - infinity

3. RHS = sec²(x)/tan(x) = [ 1/cos²(x) ]*[cos(x) / sin(x)] .... (when you divide by tan(x), the cos and sin flip..)

If you simplify, you get 1/[cos(x)sin(x)] = LHS

4. Note that equation you wrote is a cubic function, general solutions are all values of x that satisfy the equation. So simply factorise it and let the factors equal to zero.

Note that it asks for general solutions, so I have a feeling that it has an irreducible quadratic function of sin(x) after the factorisation.


Hope that helped

 

[SoulSearcher]

1) d²y/dx² = 9sinx3x
dy/dx = -3cos3x + c, where c is some constant and as dy/dx = 0 at a stationary point,
0 = -3cos(3pi/2) + c
0 = 0 + c
c = 0, therefore dy/dx = -3cos3x
therefore y = -sin3x + k, where k is some constant and as when x = pi/2, y = 1
1 = -sin(3pi/2) + k
1 = 1 + k
k = 0
therefore y = -sin3x

 

[hyparzero]

for some questions, its best to just give the procedures needed to do them, otherwise, it becomes a reliance on others, which doesnt help at all to your overall study of maths.