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(sin 70)^2-(sin 10)^2=(root 3)/2sin 80 prove 

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taliaxe | Student, College Freshman | (Level 1) eNoter

Posted May 19, 2013 at 4:29 PM via web

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(sin 70)^2-(sin 10)^2=(root 3)/2sin 80


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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted May 19, 2013 at 4:46 PM (Answer #1)

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You need to test if the given expression is a trigonometric identity, hence, you may start by converting the difference of squares to the left, into a product, such that:

`sin^2 70^o - sin^2 10^o = (sin 70^o - sin 10^o)(sin 70^o + sin 10^o)`

You need to convert the difference and sum of sines into a product, using the following formulas, such that:

`sin a - sin b = 2(sin((a-b)/2))(cos((a+b)/2))`

`sin a + sin b = 2(cos((a-b)/2))(sin((a+b)/2))`

Reasoning by analogy yields:

`sin 70^o - sin 10^o = 2(sin((70^o-10^o)/2))(cos((70^o +10^o)/2))`

`sin 70^o - sin 10^o = 2(sin(30^o))(cos(40^o))`

`sin 70^o + sin 10^o = 2(cos(30^o))(sin(40^o))`

Replacing `2(sin(30^o))(cos(40^o))*2(cos(30^o))(sin(40^o))` for the difference of squares `sin^2 70^o - sin^2 10^o` yields:

`2(sin(30^o))(cos(40^o))*2(cos(30^o))(sin(40^o)) = sqrt3/2*sin80^o`

Re-arranging the factors yields:

`2(sin(30^o))(cos(30^o))*2(sin(40^o))(cos(40^o)) =sqrt3/2*sin80^o`

Using the double angle identity yields:

`sin(2*30^o)*sin(2*40^o) = sqrt3/2*sin80^o`

`sin 60^o*sin 80^o = sqrt3/2*sin80^o`

Replacing ` sqrt3/2` for `sin 60^o` yields:

`sqrt3/2*sin80^o = sqrt3/2*sin80^o`

Hence, testing if the given identity is valid yields that` sin^2 70^o - sin^2 10^o = sqrt3/2*sin80^o` holds.

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