Recent content by chris99191

Unsure if this proof is correct I completed this proof in the attachment but can someone please check my working The aim was to show that the height of P above floor after the crate is tilted is h(cosb+2sinb) From the rectangles you can work out Length of OP=h√5 sin(a)=1/√5 cos(a)=2/√5...

I completed this proof in the attachment but can someone please check my working The aim was to show that the height of P above floor after the crate is tilted is h(cosb+2sinb) From the rectangles you can work out Length of OP=h√5 sin(a)=1/√5 cos(a)=2/√5 therefore...

i feel stupid for asking this but what subangles should i now find

That sounds pretty good haha except i don't understand where the z-shape is. Is it PA,AB and BO? and what do you mean by subangles

This still doesn't help me understand sorry. I think we have to put them into the sin general solution formula but I am not sure how

[b]1. For the diagram in the attachment, prove that the height of P above floor after being tilted is h(cosb+2sinb) [b]2. h(cosb+2sinb) [b]3. I think you need to divide them up into triangles and then use the angles for each as well as maybe using a sums to products formula

[b]1. A rectangle crate has height, h, which is half the length 2h. O is the bottom left corner and P is the top right. When a line is drawn through O to P it makes the angle a. When the crate is tilted on O it makes angle b with the ground. Show that the height of P above floor after being...

[b]1. First one is (sin2x+sinx)/(cos2x+cosx+1)=tanx Second one is (sec∂-tan∂)²=(1-sin∂)/(1+sin∂) [b]2. Sec=1/cos tan=sin/cos cos²x+sin²x=1 [b]3. 1. I think eventually the sinx/cosx need to cancel to make tanx and the 1 could be used to create a lot of options 2. I have tried to...

[b]1. Find the general solution to sin3x+sin2x=0 [b]2. I am not sure what to use. I think the sin2x=2sinxcosx but I am unsure where to go from there [b]3. sin3x+2sinxcosx=0 unsure