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

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madmoni06 | Student | Honors

Posted April 6, 2013 at 8:55 PM via web

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

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lfryerda | High School Teacher | (Level 2) Educator

Posted April 6, 2013 at 10:28 PM (Answer #2)

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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` .

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pramodpandey | College Teacher | Valedictorian

Posted April 7, 2013 at 10:11 AM (Answer #3)

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`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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