Substituting 361 in place of x,
f(361+√361)=361–√361
⇒f(361+19)=361–19
⇒f(380)=342
Substituting 361 in place of x,
f(361+√361)=361–√361
⇒f(361+19)=361–19
⇒f(380)=342
tr=14r2–1
=1(2r–1)(2r+1)=12(12r–1–12r+1)
Sn=n∑r=112(12r–1–12r+1)
Sn=12[(11–13)+(13–15)+(15–17)+…….+(12n–1–12n+1)]
Sn=12[1–12n+1]=n2n+1
The given expression can be written as,
12[(1–cos2∘)+(1–cos4∘)+…………….+(1–cos360∘)]
=12[(1+1+……..+1)–(cos2∘+cos4∘+………..+cos360∘)]
There are 180 terms in the series.
=12[180–cos(2∘+180–12×2∘)sin(180×2∘2)sin(2∘2)]
=12(180–cos181∘sin180∘sin1∘)=90
LHS = [(1+sinθ+icosθ)(sinθ+icosθ){1+(sinθ–icosθ)}(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(π2–θ)+isin(π2–θ)}n
=cosn(π2–θ)+isinn(π2–θ)
= RHS
We have, e–1=1+(–1)+(–1)22!+(–1)33!+(–1)44!+……….
The given integral = ∫(e–1)xdx=∫e–xdx=–e–x+C
L.H.S.=cosπ7–cos2π7+cos3π7
= cosπ7+cos3π7–cos(π–5π7)
= cosπ7+cos3π7+cos5π7
= cos(π7+3–12.2π7).sin(3.2π/72)sin(2π/72) Continue reading Prove That,
cosπ7−cos2π7+cos3π7=12
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 x∈R
(6x+5)2=353x2+5x+2