Specialist 1/2 Question Thread!

[Syndicate]

I am still a bit confused did you use the double angle formula for the 3rd line? Also where did the cos^2theta - sin^theta come from in line 4?


How would you prove q19?

sin4x can be simplified down to 2sin2xcos2x.


Here we use the sin addition formula.

since sin2xcos2x is the same as cos2xsin2x, you are able to add them together.



In the fourth line, Rui has used the trigonometric double angle formulas.




Q19)




- Expand this using the double angle formula.



- Now we apply the double angle formula once again.


- Expand (2cos^2(x) - 1)^2 (Note (2cos^2(x)-1)^2 = (2cos^2(x)-1)(2cos^2(x) -1) and can be expanded using the rule, (a+b)^2 = a^2 = 2ab + b^2)






............... QED

[anotherworld2b]

thank you so much  syndicate

[anotherworld2b]

Could i get further explanation please? I am still quite the novice in how to approach this question. Would you use the formula for sin (a+x)?

This is a fairly routine method, outlined in the Cambridge textbook (SM 1/2)



Then solve on CAS. You'll find this method works as long as sin and cos have the same argument.

[Syndicate]

Could i get further explanation please? I am still quite the novice in how to approach this question. Would you use the formula for sin (a+x)?

Simlifying 2sin(x) + 5cos(x) = 3 to rcos(x-A) or rsin(x+B) will yield the same result. However, since sin(x)sin(u) + cos(x)cos(u) can easily be simplified down to cos(x-u), it would be more relevant to convert it to cos(x-u) than to sin(x+u) (as then, it would require more working out).

[Gogo14]

In a lift that is accelerating upwards at 2 m/s2, a spring balance shows the apparent
weight of an object to be 2.5 kg wt. What would be the reading if the lift were at rest?

[Syndicate]

In a lift that is accelerating upwards at 2 m/s2, a spring balance shows the apparent
weight of an object to be 2.5 kg wt. What would be the reading if the lift were at rest?

2.5 = m(9.8+2)

m = 0.211864406 (try to get as many decimal places possible)

scale reading = m x (g+0)
= 2.1 kg wt.

[Jakeybaby]

2.5 = m(9.8+2)

m = 0.211864406 (try to get as many decimal places possible)

scale reading = m x (g+0)
= 2.1 kg wt.

I'd personally leave your value of m as a fraction, this would allow for more accurate results without having to waste time writing out 300 decimal places.

\begin{equation}{\frac{25}{118}}~.~{\left(9.81+0 \right)}\end{equation}

[Gogo14]

2.5 = m(9.8+2)

m = 0.211864406 (try to get as many decimal places possible)

scale reading = m x (g+0)
= 2.1 kg wt.

Why is it (9.8+2)? So lost

[jamonwindeyer]

Why is it (9.8+2)? So lost

The formula used is \(W=mg\), with that additional 2\(ms^{-1}\) coming from the upwards acceleration of the lift! In the lifts frame of reference, this is experienced as a 2\(ms^{-1}\) acceleration downwards (think of being pushed to the floor on a rollercoaster that accelerates upwards, this is the same thing)

[Adequace]

For this question http://m.imgur.com/3UDbKU3

So I found N = 8cos25, but the answer is resolving the normal force in to its components and then equating the vertical component to the weight. Isn't the reaction of the plane to the body, the entire normal force? They got 8/cos25

[Syndicate]

For this question http://m.imgur.com/3UDbKU3

So I found N = 8cos25, but the answer is resolving the normal force in to its components and then equating the vertical component to the weight. Isn't the reaction of the plane to the body, the entire normal force? They got 8/cos25

The answer given by the book is correct, as the Force F is perpendicular to the weight force not the normal force. Therefore, normal force = 8/ cos25.

(One of the ways to approach such a question is to draw a triangle (with all the forces). )

[Gogo14]

A stone is projected upwards with a speed of 14 ms from a point O at the top of a mineshaft. Five seconds earlier, a lift began to descend the mineshaft from O with a constant speed of 3.5ms. Find the depth of the lift to the nearest metre at the instant when the stone falls on it. Neglect air resist
Ans: 33m
I had a look at the worked solutions, but that only made me more confused

[ezferns]

Hey, anyone know how to derive Lami's Rule using the Sine Rule and a triangle?
I can't really picture it in my head
This is for statics btw

[Gogo14]

http://imgur.com/dhCbbsi
How do you do question 16b?
Ans for a is ]100/(9g)
Ans for b is ]2sqrt3

[Syndicate]

http://imgur.com/dhCbbsi
How do you do question 16b?
Ans for a is ]100/(9g)
Ans for b is ]2sqrt3

You can solve this question using the work-energy principle (I am not sure if it is part of the specialist curriculum).