Question

is sin^2 π/6 , sin^2 π/4 , sin^2 π/3 in AP

Answer 1

$\huge\boxed{\underline{\underline{\red{\tt question—>}}}}$

is $\tt sin^2(π/6) , sin^2(π/4) , sin^2(π/3)$in A.P?

$\huge\boxed{\underline{\underline{\blue{\tt solution—}}}}$

to prove the above statement in A.P, we can show that the middle term is the average of the other 2 terms

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we can say that:

$\tt sin^2(π/4) = \dfrac{sin^2(π/6)+sin^2(π/3)}{2}$

$\tt ==> sin^2(45°)=\dfrac{sin^2(30°)+sin^2(60°)}{2}$

$\tt ==> \dfrac{(1)^2}{(✓2)^2}=\dfrac{\dfrac{(1)^2}{(2)^2}+\dfrac{(✓3)^2}{(2^2)}}{2}$

the values used above are:

sin(30)=1/2

sin(45)=1/√2

sin(60)=√3/2

$\tt ==> \dfrac{1}{2} = \dfrac{\dfrac{1}{4} + \dfrac{3}{4}}{2}$

$\tt ==> \dfrac{1}{2} = \dfrac{1}{2}$

since at this point on this equation L.H.S=R.H.S holds up and it is in this position that we can conclude that the above series are in A.P

hope my answer helps you out!!