Mathematics: Post your doubts here!

[whitecorp]

Hi. Can anyone solve this question for me . Iwould be grateful.

(i)Prove The Identity cos4x + 4cos2x = 8(cosx)^2 -3
(ii) Hence solve the equations for 0<x<360.

plz solve it with detail showing every important step. Thanks in advance.

Your identity is incorrect, the RHS should be 8 (cosx)^4 -3.
I have provided the solution for this:



For (ii), what equation exactly are you referring to? Note that (i) is an identity, there is nothing to solve for.

Peace.

 

[GorgeousEyes]

There is symmetry involved here-the path of the ball on the way upwards, and that of the ball on the way downwards.
Let initial velocity of the ball be u, then time taken to travel upwards =3/2 =1.5s
When it attains its maximum height before falling downwards, velocity =0
Hence, 0= u + (-g)(1.5) =======> u=1.5g m/s (shown)
(Note that I have assigned upwards as being positive for the first half of the journey, therefore acceleration due to free fall aka g is defined in the negative sense.)
Also, 0^2 = u^2 -2(g)(s)
Substituting in (1): 0=2.25g^2 - 2g(s)
Solving gives s = 2.25/2 =1.125 g metres. (shown)

(Not sure if you use g =9.81 or 10 m/s^2, so I will leave it to you to stick in the appropriate value for both answers)

Hope this helps. Peace.

Thanks a millioon , Helped me a looot

 

[whitecorp]

Thanks a millioon , Helped me a looot

No problem. Peace.

 

[Latifa]

Hi
how can i find IGCSE 2012 pastpapers for chem,physics and maths???
can any one help me p/zzzz!???
i need them urgentlyyy!!!

 

[asexamskillme111]

3^(x+2)=3^x+3^2

anyone?

 

[whitecorp]

3^(x+2)=3^x+3^2

anyone?

Consider the substitution y= 3^x. Hope this helps. Peace.

 

[kronix6]

Your identity is incorrect, the RHS should be 8 (cosx)^4 -3.
I have provided the solution for this:



For (ii), what equation exactly are you referring to? Note that (i) is an identity, there is nothing to solve for.

Peace.

THANK U SO MUCH> I APPRECIATE IT ALOT> ILIKED UR FB PAGE

 

[whitecorp]

THANK U SO MUCH> I APPRECIATE IT ALOT> ILIKED UR FB PAGE

Appreciate the gesture, peace.

 

[bloody_mary]

Did you mean to say one fourth?

and you really didn't know how to answer the questions because of a typo? -_-

thanks anyways

 

[whitecorp]

and you really didn't know how to answer the questions because of a typo? -_-

thanks anyways

Oh no, I can't, my math is awful, terrible. Peace.

 

[VelaneDeBeaute]

Okay, so there was a question of Trigonometry.
If arctan A = x and arctan B = y, express tan(A+B) in terms of x and y.
The answer is x + y / 1 - xy. I'm missing a tiny concept but I dunno where ofcourse.
XPFMember smzimran OakMoon! whitecorp BadRobot14

 

[XPFMember]

Okay, so there was a question of Trigonometry.
If arctan A = x and arctan B = y, express tan(A+B) in terms of x and y.
The answer is x + y / 1 - xy. I'm missing a tiny concept but I dunno where ofcourse.
XPFMember smzimran OakMoon! whitecorp BadRobot14

AsSalamoAlaikum Wr Wb

Alright, let me check..

 

[XPFMember]

Okay, so there was a question of Trigonometry.
If arctan A = x and arctan B = y, express tan(A+B) in terms of x and y.
The answer is x + y / 1 - xy. I'm missing a tiny concept but I dunno where ofcourse.
XPFMember smzimran OakMoon! whitecorp BadRobot14

AsSalamoAlaikum Wr Wb


tan (A + B) = (tan A + tan B)/(1 - (tan A)(tan B))

from the info given, rearrange, and you get: x = tan A ; y = tan B

just substitute, and you get x + y / 1 - xy

 

[bamteck]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w09_qp_42.pdf

whitecorp please help me for no. 6 (ii) & (iii)
Thanks

 

[whitecorp]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_42.pdf

whitecorp please help me for no. 6 (ii) & (iii)
Thanks

For (i) I obtained an acceleration value of 4 m/s^2.

(ii) Maximum height reached by A =distance B falls through before reaching the ground
Let this be s.
Then using v^2 =u^2 + 2as, substituting in the values, we have
1.6^2 = 0^2 +2 (4)(s) =====> s= 0.32 m (shown)

(iii) Using s=ut + 0.5*a*t^2, substituting in the values, we have
0.32 = 0.5*(4)*t^2 =2 t^2
Hence, t^2 = 0.16 =====> t=0.4 s (shown)

Hope this helps. Peace.

 

[bamteck]

For (i) I obtained an acceleration value of 4 m/s^2.

(ii) Maximum height reached by A =distance B falls through before reaching the ground
Let this be s.
Then using v^2 =u^2 + 2as, substituting in the values, we have
1.6^2 = 0^2 +2 (4)(s) =====> s= 0.32 m (shown)

(iii) Using s=ut + 0.5*a*t^2, substituting in the values, we have
0.32 = 0.5*(4)*t^2 =2 t^2
Hence, t^2 = 0.16 =====> t=0.4 s (shown)

Hope this helps. Peace.

Yeah I agree with you !
But the answer isn't right I guess
Have a look in the marking scheme !
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w09_ms_42.pdf

 

[leadingguy]

the way i attemptd the qstn is totally different can anyone explain me how this to be done???? the highlighted part. down is the ms of the qstn as well thanks

 

[iKhaled]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_42.pdf

whitecorp please help me for no. 6 (ii) & (iii)
Thanks

Yeah I agree with you !
But the answer isn't right I guess
Have a look in the marking scheme !
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_ms_42.pdf

(ii)the distance moved by B is,
1.6^2 = 0+2(4)s
s= 0.32 m

this is NOT the maximum height reached by A, this is only the the distance A moved above the ground while b was moving downard. from this instant B will be touchin the ground so there will be no tension in the string anymore because the string is slack which means from now on A is moving upward under gravity (g ms^-2 ). so the maximum height A has moved is the distance it moved while b was moving downward PLUS the distance it moved till it reaches 0 velocity ( max height )
V^2 = u^2 + 2as
0 = 1.6^2 + 2(-10)(s)
s = 0.128

total distance moved by A = 0.32 + 0.128 = 0.448m

(iii) now for the total time taken, in the first stage when it was moving with an acceleration of 4ms^-2 it was:

v = u + at
1.6 = o + 4t
t= 0.4 s

now for the second stage when it was moving with a DECELERATION of 10 ms^-2

v = u + at
0 = 1.6 + -10t
t= 0.16 s

total time taken = 0.4 + 0.16
total time taken = 0.56 s

hope this makes sense to u, anymore questions feel free to ask me

 

[bamteck]

(ii)the distance moved by B is,
1.6^2 = 0+2(4)s
s= 0.32 m

this is NOT the maximum height reached by A, this is only the the distance A moved above the ground while b was moving downard. from this instant B will be touchin the ground so there will be no tension in the string anymore because the string is slack which means from now on A is moving upward under gravity (g ms^-2 ). so the maximum height A has moved is the distance it moved while b was moving downward PLUS the distance it moved till it reaches 0 velocity ( max height )
V^2 = u^2 + 2as
0 = 1.6^2 + 2(-10)(s)
s = 0.128

total distance moved by A = 0.32 + 0.128 = 0.448m

(iii) now for the total time taken, in the first stage when it was moving with an acceleration of 4ms^-2 it was:

v = u + at
1.6 = o + 4t
t= 0.4 s

now for the second stage when it was moving with a DECELERATION of 10 ms^-2

v = u + at
0 = 1.6 + -10t
t= 0.16 s

total time taken = 0.4 + 0.16
total time taken = 0.56 s

hope this makes sense to u, anymore questions feel free to ask me

Yeah thank you very much !
I'll pm you ? Ok ?

 

[iKhaled]

Yeah thank you very much !
I'll pm you ? Ok ?

no problem and yeah ok, feel free to !