If sinα, sin^2α and 1 are in A.P, (0< or = α < or = 90°), Find α

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An arithmetic progression is a sequence of numbers that have a common difference between each term in the sequence. This means that `sin^2a-sina` , and `1-sin^2a` must be the same difference, which gives the equation:

`sin^2a-sina=1-sin^2a` where a is acute

`2sin^2a-sina-1=0` factor this

`(2sina+1)(sina-1)=0`

This means that either `sina=-1/2` or `sina=1` . Since a is between 0 and 90 degrees, then the first equation is not valid, so `a=pi/2` .

**The value of a is `a=pi/2` .**

`sin(a),sin^2(a),1 ` are in Arithmetic progression. Therefore

`2sin^2(a)=1+sin(a)`

`2sin^2(a)-sin(a)-1=0`

Let `sin^2(a)=x` ,therefore above equation reduces to

`2x^2-x-1=0`

factoring by spliting middle term.

`2x^2-2x+x-1=0`

`2x(x-1)+1(x-1)=0`

`(x-1)(2x+1)=0`

either x-1=0

or 2x+1=0

if x-1=0

x=1

`sin(a)=1`

`a=pi/2`

so AP will be 1,1,1

if 2x+1=0

2x=-1

x=-1/2

then `sin(a)=-1/2`

`a=-pi/6`

so A.P. will be -1/2,1/4,1

So

`a=-pi/6 or pi/2`

**In above answer a=pi/2**

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