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Demonstrate that 1/sin 15 degres - 1/cos 15 degrees= 2 `sqrt2`

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minlux | Honors

Posted July 25, 2013 at 3:12 PM via web

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Demonstrate that 1/sin 15 degres - 1/cos 15 degrees= 2 `sqrt2`

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted July 25, 2013 at 3:42 PM (Answer #1)

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You need to bring the fractions to the left to a common denominator, such that:

`(cos 15^o - sin15^o)/(sin 15^o*cos 15^o) = 2sqrt2`

You should notice that if denominator would be multiplied by 2, the double angle identity can be used, hence, you need to multiplicate by `1/2` both sides, such that:

`(cos 15^o - sin15^o)/(2sin 15^o*cos 15^o) = (1/2)*2sqrt2`

Using the double angle identity `2 sin alpha*cos alpha = sin 2 alpha ` yields:

`(cos 15^o - sin15^o)/(sin 2*15^o) = sqrt 2`

`(cos 15^o - sin15^o)/(sin 30^o) = sqrt 2`

Since `sin 30^o = 1/2` , yields:

`(cos 15^o - sin15^o)/(1/2) = sqrt 2`

`2(cos 15^o - sin15^o) = sqrt 2 => cos 15^o - sin15^o = sqrt2/2`

You should use the following identity, such that:

`cos 15^o = sin(90^o - 15^o) = sin 75^o`

Replacing `sin 75^o` for `cos 15^o` yields:

`sin 75^o - sin15^o = sqrt2/2`

Converting the difference into a product yields:

`2cos((75^o + 15^o)/2)sin((75^o - 15^o)/2) = sqrt2/2`

`2cos(45^o)*sin(30^o) = sqrt2/2`

Replacing `sqrt2/2` for `cos(45^o)` and `1/2` for `sin(30^o)` yields:

`2*(sqrt2/2)*(1/2) = sqrt2/2`

Reducing duplicate factors to the left, yields:

`sqrt2/2 = sqrt2/2`

Hence, performing the indicated conversions, yields that the given identity holds.

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