What did Brahmo samaj preach?
Answers
Answer:
Brahmo Samaj believed in the fatherhood of God and brotherhood of mankind. It not only discarded meaningless rites and rituals but also forbade idol-worship. Raja Ram Mohan Roy believed in the oneness of God and preached the same. It preached to love human beings, stop animal sacrifice and offerings.
Explanation:
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Answer:
Answer
³/₂
Formula Used
\begin{gathered}\rm cos2\theta+1=2cos^2\theta\\\\\bullet \;\; \rm cos^2\theta=\dfrac{cos2\theta+1}{2}\\\\\bullet \;\; \rm cosA+cosB=\tiny{2cos(\dfrac{A+B}{2}).cos(\dfrac{A-B}{2})}\\\\\bullet \;\; \rm cos(\pi-\theta)=-sin\theta\end{gathered}
cos2θ+1=2cos
2
θ
∙cos
2
θ=
2
cos2θ+1
∙cosA+cosB=2cos(
2
A+B
).cos(
2
A−B
)
∙cos(π−θ)=−sinθ
Solution
LHS
\begin{gathered}\rm \to cos^2x+cos^2(x+\dfrac{\pi}{3})+cos^2(x-\dfrac{\pi}{3})\\\\\end{gathered}
→cos
cos2θ+1=2cos
2
θ
∙cos
2
θ=
2
cos2θ+1
∙cosA+cosB=2cos(
2
A+B
).cos(
2
A−B
)
∙cos(π−θ)=−sinθ
Solution
LHS
\begin{gathered}\rm \to cos^2x+cos^2(x+\dfrac{\pi}{3})+cos^2(x-\dfrac{\pi}{3})\\\\\end{gathered}
→cos
2
x+cos
2
(x+
3
π
)+cos
2
(x−
3
π
)
\begin{gathered}\rm \to \dfrac{cos2x+1}{2}+\dfrac{\tiny{cos2(x+\dfrac{\pi}{3})+1}}{2}+\dfrac{\tiny{cos2(x-\dfrac{\pi}{3})+1}}{2}\\\\\to \rm \dfrac{1}{2}\tiny{( 3+cos2x+cos(2x+\dfrac{2\pi}{3})+cos(2x-\dfrac{2\pi}{3}))}\\\\\to \rm \dfrac{1}{2} \tiny{(3+cos2x+2cos(\dfrac{2x+\dfrac{2\pi}{3}+2x-\dfrac{2\pi}{3}}{2}).cos(\dfrac{2x+\dfrac{2\pi}{3}-2x+\dfrac{2\pi}{3}}{2}))}\\\\\end{gathered}
→
2
cos2x+1
+
2
cos2(x+
3
π
)+1
+
2
cos2(x−
3
π
)+1
→
2
1
(3+cos2x+cos(2x+
3
2π
)+cos(2x−
3
2π
))
→
2
1
(3+cos2x+2cos(
2
2x+
3
2π
+2x−
3
2π
).cos(
2
2x+
3
2π
−2x+
3
2π
))
\begin{gathered}\rm \to \dfrac{1}{2}(3+cos2x+2cos2x.(-sin30^0))\\\\\rm \to \dfrac{1}{2}(3+cos2x+2cos2x.\dfrac{-1}{2})\\\\\rm \to \dfrac{1}{2}(3+cos2x-cos2x)\\\\\rm \to \dfrac{3}{2}\\\\\rm \to RHS\end{gathered}
→
2
1
(3+cos2x+2cos2x.(−sin30
0
))
→
2
1
(3+cos2x+2cos2x.
2
−1
)
→
2
1
(3+cos2x−cos2x)
→
2
3
→RHS
Since , LHS = RHS ,
Hence proved
Explanation:
Answer
³/₂
Formula Used
\begin{gathered}\rm cos2\theta+1=2cos^2\theta\\\\\bullet \;\; \rm cos^2\theta=\dfrac{cos2\theta+1}{2}\\\\\bullet \;\; \rm cosA+cosB=\tiny{2cos(\dfrac{A+B}{2}).cos(\dfrac{A-B}{2})}\\\\\bullet \;\; \rm cos(\pi-\theta)=-sin\theta\end{gathered}
cos2θ+1=2cos
2
θ
∙cos
2
θ=
2
cos2θ+1
∙cosA+cosB=2cos(
2
A+B
).cos(
2
A−B
)
∙cos(π−θ)=−sinθ
Solution
LHS
\begin{gathered}\rm \to cos^2x+cos^2(x+\dfrac{\pi}{3})+cos^2(x-\dfrac{\pi}{3})\\\\\end{gathered}
→cos
2
x+cos
2
(x+
3
π
)+cos
2
(x−
3
π
)
\begin{gathered}\rm \to \dfrac{cos2x+1}{2}+\dfrac{\tiny{cos2(x+\dfrac{\pi}{3})+1}}{2}+\dfrac{\tiny{cos2(x-\dfrac{\pi}{3})+1}}{2}\\\\\to \rm \dfrac{1}{2}\tiny{( 3+cos2x+cos(2x+\dfrac{2\pi}{3})+cos(2x-\dfrac{2\pi}{3}))}\\\\\to \rm \dfrac{1}{2} \tiny{(3+cos2x+2cos(\dfrac{2x+\dfrac{2\pi}{3}+2x-\dfrac{2\pi}{3}}{2}).cos(\dfrac{2x+\dfrac{2\pi}{3}-2x+\dfrac{2\pi}{3}}{2}))}\\\\\end{gathered}
→
2
cos2x+1
+
2
cos2(x+
3
π
)+1
+
2
cos2(x−
3
π
)+1
→
2
1
(3+cos2x+cos(2x+
3
2π
)+cos(2x−
3
2π
))
→
2
1
(3+cos2x+2cos(
2
2x+
3
2π
+2x−
3
2π
).cos(
2
2x+
3
2π
−2x+
3
2π
))
\begin{gathered}\rm \to \dfrac{1}{2}(3+cos2x+2cos2x.(-sin30^0))\\\\\rm \to \dfrac{1}{2}(3+cos2x+2cos2x.\dfrac{-1}{2})\\\\\rm \to \dfrac{1}{2}(3+cos2x-cos2x)\\\\\rm \to \dfrac{3}{2}\\\\\rm \to RHS\end{gathered}
→
2
1
(3+cos2x+2cos2x.(−sin30
0
))
→
2
1
(3+cos2x+2cos2x.
2
−1
)
→
2
1
(3+cos2x−cos2x)
→
2
3
→RHS
Since , LHS = RHS ,
Hence provedAnswer
³/₂
Formula Used
\begin{gathered}\rm cos2\theta+1=2cos^2\theta\\\\\bullet \;\; \rm cos^2\theta=\dfrac{cos2\theta+1}{2}\\\\\bullet \;\; \rm cosA+cosB=\tiny{2cos(\dfrac{A+B}{2}).cos(\dfrac{A-B}{2})}\\\\\bullet \;\; \rm cos(\pi-\theta)=-sin\theta\end{gathered}
cos2θ+1=2cos
2
θ
∙cos
2
θ=
2
cos2θ+1
∙cosA+cosB=2cos(
2
A+B
).cos(
2
A−B
)
∙cos(π−θ)=−sinθ
Solution
LHS
\begin{gathered}\rm \to cos^2x+cos^2(x+\dfrac{\pi}{3})+cos^2(x-\dfrac{\pi}{3})\\\\\end{gathered}
→cos
2
x+cos
2
(x+
3
π
)+cos
2
(x−
3
π
)
\begin{gathered}\rm \to \dfrac{cos2x+1}{2}+\dfrac{\tiny{cos2(x+\dfrac{\pi}{3})+1}}{2}+\dfrac{\tiny{cos2(x-\dfrac{\pi}{3})+1}}{2}\\\\\to \rm \dfrac{1}{2}\tiny{( 3+cos2x+cos(2x+\dfrac{2\pi}{3})+cos(2x-\dfrac{2\pi}{3}))}\\\\\to \rm \dfrac{1}{2} \tiny{(3+cos2x+2cos(\dfrac{2x+\dfrac{2\pi}{3}+2x-\dfrac{2\pi}{3}}{2}).cos(\dfrac{2x+\dfrac{2\pi}{3}-2x+\dfrac{2\pi}{3}}{2}))}\\\\\end{gathered}
→
2
cos2x+1
+
2
cos2(x+
3
π
)+1
+
2
cos2(x−
3
π
)+1
→
2
1
(3+cos2x+cos(2x+
3
2π
)+cos(2x−
3
2π
))
→
2
1
(3+cos2x+2cos(
2
2x+
3
2π
+2x−
3
2π
).cos(
2
2x+
3
2π
−2x+
3
2π
))
\begin{gathered}\rm \to \dfrac{1}{2}(3+cos2x+2cos2x.(-sin30^0))\\\\\rm \to \dfrac{1}{2}(3+cos2x+2cos2x.\dfrac{-1}{2})\\\\\rm \to \dfrac{1}{2}(3+cos2x-cos2x)\\\\\rm \to \dfrac{3}{2}\\\\\rm \to RHS\end{gathered}
→
2
1
(3+cos2x+2cos2x.(−sin30
0
))
→
2
1
(3+cos2x+2cos2x.
2
−1
)
→
2
1
(3+cos2x−cos2x)
→
2
3
→RHS
Since , LHS = RHS ,
Hence proved