Category Archives: Math

Sn=13+115+135+163+=?

tr=14r21

=1(2r1)(2r+1)=12(12r112r+1)

Sn=nr=112(12r112r+1)

Sn=12[(1113)+(1315)+(1517)+.+(12n112n+1)]

Sn=12[112n+1]=n2n+1

sin21+sin22++sin2180=?

The given expression can be written as,

12[(1cos2)+(1cos4)+.+(1cos360)]

=12[(1+1+..+1)(cos2+cos4+..+cos360)]

There are 180 terms in the series.

=12[180cos(2+18012×2)sin(180×22)sin(22)]

=12(180cos181sin180sin1)=90

Prove that, (1+sinθ+icosθ1+sinθicosθ)n=cosn(π2θ)+isinn(π2θ)

LHS = [(1+sinθ+icosθ)(sinθ+icosθ){1+(sinθicosθ)}(sinθ+icosθ)]n

=[(1+sinθ+icosθ)(sinθ+icosθ)(sinθ+icosθ)+{sin2θ(icosθ)2}]n

=[(1+sinθ+icosθ)(sinθ+icosθ)(sinθ+icosθ)+(sin2θ+cos2θ)]n

=[(1+sinθ+icosθ)(sinθ+icosθ)(1+sinθ+icosθ)]n

=(sinθ+icosθ)n

={cos(π2θ)+isin(π2θ)}n

=cosn(π2θ)+isinn(π2θ)

= RHS

Integration of Infinite Series

(12!13!+14!15!+)xdx=?

We have, e1=1+(1)+(1)22!+(1)33!+(1)44!+.

The given integral = (e1)xdx=exdx=ex+C

Prove That,

cosπ7cos2π7+cos3π7=12

L.H.S.=cosπ7cos2π7+cos3π7

= cosπ7+cos3π7cos(π5π7)

= cosπ7+cos3π7+cos5π7

= cos(π7+312.2π7).sin(3.2π/72)sin(2π/72) Continue reading Prove That,

cosπ7cos2π7+cos3π7=12

Solve for xR

(6x+5)2=353x2+5x+2

We have, 36x2+60x+25=353x2+5x+2

12(3x2+5x)+25=35(3x2+5x)+2

Let, 3x2+5x=t

So, 12t+25=35t+2

12t2+49t+15=0 Continue reading Solve for xR

(6x+5)2=353x2+5x+2