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Mathematics: Post your doubts here!

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NOV 2008 P4
Q7 A particle P is held at rest at a fixed point O and then released. P falls freely under gravity until it reaches the point A which is 1.25m below O.
(i) Find the speed of P at A and the time taken for P to reach A.[3]
The particle continues to fall, but now its downward acceleration t seconds after passing through A is (10−0.3t)ms−2.
(ii) Find the total distance P has fallen, 3s after being released from O............................
 
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Can someone please do this
View attachment 51697
PART 1-
sin (a+b) =sinacosb+cosasinb
*so for sin (3x) = = sin(2x + x)

= sin(2x)cos(x) + cos(2x)sin(x)
=cosx [sin(2x)cos(x) + cos(2x)sin(x)]
= 2sin(x)cos(x)cos(x) + (1 - 2sin² (x) ) sin(x)
= 2sin(x)cos² (x) + sin(x) - 2sin³ (x)
= 2sin(x) ( 1 -sin² (x) ) + sin(x) - 2sin³ (x)
= 2sin(x) -2sin³ (x) + sin(x) - 2sin³ (x)
= -4sin³ (x) + 3sin(x)
= 3sin(x) - 4sin³ (x)
 
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Hi,
I'm doing trigonometry P1 and I'm stuck at Exercise 10B : Use the symmetric and periodic properties of the sine, cosine and tangent functions to establish the following results:
(a.) sin(90-theta) = cos theta
(b.) sin(270+theta) = -cos theta
(c.) sin(90+theta) = cos theta
(d.) cos(90+theta) = -sin theta
(e.) tan(theta - 180) = tan theta
How do I solve these? Please help.
 
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Hi,
I'm doing trigonometry P1 and I'm stuck at Exercise 10B : Use the symmetric and periodic properties of the sine, cosine and tangent functions to establish the following results:
(a.) sin(90-theta) = cos theta
(b.) sin(270+theta) = -cos theta
(c.) sin(90+theta) = cos theta
(d.) cos(90+theta) = -sin theta
(e.) tan(theta - 180) = tan theta
How do I solve these? Please help.
which year is this? Just interested.
 
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Hi,
I'm doing trigonometry P1 and I'm stuck at Exercise 10B : Use the symmetric and periodic properties of the sine, cosine and tangent functions to establish the following results:
(a.) sin(90-theta) = cos theta
(b.) sin(270+theta) = -cos theta
(c.) sin(90+theta) = cos theta
(d.) cos(90+theta) = -sin theta
(e.) tan(theta - 180) = tan theta
How do I solve these? Please help.
Remember if the digits are odd, remove sin and and cos and vice versa, in case of tan, remove tan and add cot and vice versa.
Remember ASTC graph.
(a) sin(90 - theta) = 90 is odd so add cos remove sin. 90 - thetha = A so cos theta.
(b) sin (270+ theta) = 270 is odd so remove sin add cos, 270 + theta = C so sin is -ve = - cos theta.
(e) tan(-(180-theta)) = 180 is even, so no changes. tan(-theta) = tan theta.
REST TRY OUT YOURSELF, if u still have doubts, we are here to solve them.
 
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