IIT JEE , AIEEE Forum - Mathematics , Algebra

sachin_gupta1991

Q. If cos(x-y), cosx, cos(x+y) are in H.P. then cosx sec(y/2)=
Q.If in a triangle PQR, sinP,sinQ,sinR are in A.P.,then prove that the altitudes are in H.P.

Q. For any odd integer n>=1 , n³- (n-1)³+.........(-1)^n-11³=

waterdemon

Hello,
Solution(1):

Cos(x-y) , Cosx , Cos(x+y) are in HP
So we have :

Cosx = 2 Cos(x-y) Cos(x+y) / Cos(x-y) + Cos(x+y)
Cosx = 2(Cos²x - Sin²y) / 2 Cosx.Cosy

In the above step expanding and solving with Formula.
Cos(A+B) and Cos(A-B)

Cosx(Cosx.Cosy) = (Cos²x - Sin²y)
Cos²x.Cosy = (Cos²x - Sin²y)
Cos²x.Cosy - Cos²x = (-Sin²y)
Cos²x (Cosy - 1) = (-Sin²y)

OR

Cos²x(1-Cosy) = Sin²y
Now we know that,

Sin²A + Cos²A = 1
So,
Sin²y + Cos²y = 1
Sin²y = 1 - Cos²y

So we have,
Cos²x(1-Cosy) = 1 - Cos²y
Cos²x(1-Cosy) = (1)²-Cos²y

In abv step applying for A²-B² = (A+B)(A-B)

Cos²x (1-Cosy) = (1+Cosy)(1-Cosy)

Cos²x = 1 + Cosy

Now we know that,
Cos2A = 2Cos²A - 1
Cos2A + 1 = 2Cos²A
so we get:

1+Cosy = 2Cos²(y/2)

Cos²x = 2Cos²(y/2)
Cos²x.[ 1/Cos²(y/2) ] = 2
Cos²x.Sec²(y/2) = 2

Taking Square root,
Cosx.Sec(y/2) =

2

Hope You find it useful.
Cheers!!!!!!!!!!!!!!

waterdemon

Solution(3):

S = [n³ + (n-1)³ + (n-2)³ + ......... + 1³] .................
- 2 [(n-1)³ + (n-3)³ + ............... + 2³]

= n²(n+1)²/4 - 2*8[(n-1/2)³ + (n-3/2)³ + ........ 1³]
= n²(n+1)²/4 - 16*[(n-1/2)³ x {(n-1)/2 + 1}³ / 4

=n²(n+1)²/4 - 4(n-1/2)³.(n-1)/2 + 1}³

On solving we get it as:

= (1/4)(n+1)²(2n-1)

Hope it is useful.
Cheers!!!!!!!!!!!!!!!!!!

waterdemon

Solution(2):
Let side opposite to angle P be "p"
Let side opposite to angle Q be "q"
Let side opposite to angle R Be "r"

Given to us:
SinP , SinQ , SinR are in A.P
Therefore,
p,q,r will also be in A.p as in a triangle the lenght of the
sides are corresponding to their angles.

Let,
SinP/p = SinQ/q = SinR/r = K .......(1)

Also,we know altitudes are
qrSinP/p , prSinQ/q , pqSinR/r

Now using equation (1) we have:

q/pr.SinQ - p/qr.SinP = q/K.pqr - p/k.pqr = q-p/K.pqr

And Similarly:
r/pq.SinR - q/pr.SinQ = r-q/K.pqr

Now we will satisfy condition of p,q,r to be in AP.
So we put
q-p = r-q
we get p,q,r in A.P

Therefore,
r/pq.SinR - q/pr.SinQ = q/pr.SinQ - p/qr.SinP

Now we get,
2(q/pr.SinQ) = r/pq.SinR + p/qr.SinP

So we have,
p/qr.SinP , q/pr.SinQ , r/pq.SinR in A.P

And

qr.SinP/p , pr.SinQ/q , pq.SinR/r in H.P
Hence answer...
altitudes are in H.P

Proved.

Hope you find it useful.
Cheers!!!!!!!!!@@@!!!!!!!!!!

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