Mathematics: Post your doubts here!

[PANDA-]

sin x = cos (90-x)
cos x = sin (90-x)

tan(0.5pi-x) is same as tan (90-x)

tan (90-x) = sin(90-x)/cos(90-x)
= cos x/ sin x
= 1/tanx ..

hope it helps

Yes! Got it. Thanks a lot.

 

[sma786]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s03_qp_1.pdf
What sin formula did they use in Q9 part iii ?!

 

[MustafaMotani]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s03_qp_1.pdf
What sin formula did they use in Q9 part iii ?!

I think u shall get AB and BC by cos rule

 

[ZainH]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s03_qp_1.pdf
What sin formula did they use in Q9 part iii ?!

You can't use sin since it's not a right angle triangle.

For finding side AB, you know that angle BOA is 60, and two of your sides are 8. It has to be an equilateral triangle meaning AB is also 8.
For finding BC you just use cos rule using triangle OBC. If you want I can solve it :c


Also, I've got finally got question :3 A star daredevil asd
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_13.pdf
Q3.

 

[ShreeyaBeatz]

9709/11/M/J/12 Question no 7(b)(i)
how do we solve that "convergent"progression??

 

[asd]

You can't use sin since it's not a right angle triangle.

For finding side AB, you know that angle BOA is 60, and two of your sides are 8. It has to be an equilateral triangle meaning AB is also 8.
For finding BC you just use cos rule using triangle OBC. If you want I can solve it :c


Also, I've got finally got question :3 A star daredevil asd
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_13.pdf
Q3.

Oh, dude I just did it yest. D
P(a,0)
Q(0,b)
now the distance PQ= [a^2 + b^2]^1/2 = 45^1/2
a^2+b^2=45.
substituting (a,0), y=-1/2 (x + a)
2y=a-x ---> (i)
Now substituting (b,0), y-b=-1/2(x-o)
2y= 2b-x ---> (ii)
a-x=2b-x
a=2b
Now plug in the value of a in the distance equation to get the values.

 

[ZainH]

9709/11/M/J/12 Question no 7(b)(i)
how do we solve that "convergent"progression??

For arithmetic progressions, a = 1st term, a+d = 2nd term, a+2d = 3rd term , and so on.
d = the difference between consecutive terms.

The first term here is 1 and the second term here is cos²x.
SO, a=1 and a+d= cos²x

We can find d from this, as we have two equations.
Since a = 1 we can write the 2nd equation as 1+d = cos²x , which gives d as cos²x-1

Now, we use the formula for sum of a number of terms in an arithmetic progression which is Sn= n/2(2a + (n-1) d ) you should learn this formula.
Since there asking for the sum of the first ten terms we take n as 10.

S10= 10/2 (2(1) + (10 -1) cos²x-1 )
= 5 (2 + (9) cos²x-1 )

If you know identities you should recognize cos²x-1 is the same as -sin²x
So we replace that.

= 5 ( 2+ (9)(-sin²x)
= 10-45sin²x

a= 10 , b = 45.
Hope you understood.

 

[Sarah22]

3a where dot product =0
10a) idk

Oops sorry.. Not the 3(a), its 3(ii)

 

[ShreeyaBeatz]

For arithmetic progressions, a = 1st term, a+d = 2nd term, a+2d = 3rd term , and so on.
d = the difference between consecutive terms.

The first term here is 1 and the second term here is cos²x.
SO, a=1 and a+d= cos²x

We can find d from this, as we have two equations.
Since a = 1 we can write the 2nd equation as 1+d = cos²x , which gives d as cos²x-1

Now, we use the formula for sum of a number of terms in an arithmetic progression which is Sn= n/2(2a + (n-1) d ) you should learn this formula.
Since there asking for the sum of the first ten terms we take n as 10.

S10= 10/2 (2(1) + (10 -1) cos²x-1 )
= 5 (2 + (9) cos²x-1 )

If you know identities you should recognize cos²x-1 is the same as -sin²x
So we replace that.

= 5 ( 2+ (9)(-sin²x)
= 10-45sin²x

a= 10 , b = 45.
Hope you understood.

I meant Qno.7 (b)(i)

 

[daredevil]

You can't use sin since it's not a right angle triangle.

For finding side AB, you know that angle BOA is 60, and two of your sides are 8. It has to be an equilateral triangle meaning AB is also 8.
For finding BC you just use cos rule using triangle OBC. If you want I can solve it :c


Also, I've got finally got question :3 A star daredevil asd
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_13.pdf
Q3.

already solved it in the previous threads -_-
seriously i'm thinking i've done every difficult question ... thankuuuuu xpf 8D

 

[ShreeyaBeatz]

u did (a)

 

[shezi1995]

You can't use sin since it's not a right angle triangle.

For finding side AB, you know that angle BOA is 60, and two of your sides are 8. It has to be an equilateral triangle meaning AB is also 8.
For finding BC you just use cos rule using triangle OBC. If you want I can solve it :c


Also, I've got finally got question :3 A star daredevil asd
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_13.pdf
Q3.

ABC is a right angled triangle. Angle subtended by diameter is always 90 degree. So angle ABC is 90. angle c will be half of (pi/3) that is (pi/6). so you just use sin and cos with the hypotenuse AC equal to 16. So no need to use cosine rule and stuff. It is just o level maths knowledge.

 

[ShreeyaBeatz]

The first two terms of a geometric progression are 1 and (1/3) tan^2 θ respectively, where 0 < θ < (1/2)π
Find the set of values of θ for which the progression is convergent

INSTANT HELP NEEDED !

 

[A star]

well guys goodluck and happy tutoring rest of them daredevil

 

[Rutzaba]

agh.. I am tensed :/ I DONT FUKIN WANNA BE ....UGHHH

its going to be ok in sha Allah... have faith in urself and the Almighty

 

[Rutzaba]

wats wrong wth u guys... i had clashes... and then i was giving accelerated... and that was wen i paniced just keep calm and say Alhamdulilah!

 

[ZainH]

ABC is a right angled triangle. Angle subtended by diameter is always 90 degree. So angle ABC is 90. angle c will be half of (pi/3) that is (pi/6). so you just use sin and cos with the hypotenuse AC equal to 16. So no need to use cosine rule and stuff. It is just o level maths knowledge.

Cosπ/6= 16/hypotenuse
Which gives the hypotenuse as 18 something. Which is wrong, it's not a right angle triangle :c

 

[ZainH]

Oh, dude I just did it yest. D
P(a,0)
Q(0,b)
now the distance PQ= [a^2 + b^2]^1/2 = 45^1/2
a^2+b^2=45.
substituting (a,0), y=-1/2 (x + a)
2y=a-x ---> (i)
Now substituting (b,0), y-b=-1/2(x-o)
2y= 2b-x ---> (ii)
a-x=2b-x
a=2b
Now plug in the value of a in the distance equation to get the values.

I understood up until , a²+b² = 45 ..

Now what are you substituting your values into?

 

[daredevil]

The first two terms of a geometric progression are 1 and (1/3) tan^2 θ respectively, where 0 < θ < (1/2)π
Find the set of values of θ for which the progression is convergent

INSTANT HELP NEEDED !

Ok r<1 then progression is convergent. So use that in here however u can cz i cant solve ryt now sorry

 

[asd]

I understood up until , a²+b² = 45 ..

Now what are you substituting your values into?

Into the equation of straight line "y-y1=m(x-x1)"