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# How calculate sin100^o+cos100^o+sin80^o+cos80^o?

minlux | Honors

Posted September 24, 2013 at 5:42 PM via web

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How calculate sin100^o+cos100^o+sin80^o+cos80^o?

Tagged with math, trigonometry

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted September 24, 2013 at 5:56 PM (Answer #1)

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You need to form the following two groups of same trigonometric functions, such that:

`(sin100^o + sin80^o) + (cos100^o + cos80^o)`

You need to convert the summations into products, such that:

`sin100^o + sin80^o = 2sin(100^o/2 + 80^o/2)*cos(100^o/2 - 80^o/2)`

`sin100^o + sin80^o = 2sin 90^o cos 10^o`

Since `sin 90^o = 1` yields:

`sin100^o + sin80^o = 2cos 10^o`

`cos100^o + cos80^o = 2cos(100^o/2 + 80^o/2)*cos(100^o/2 - 80^o/2)`

`cos100^o + cos80^o = 2cos90^o*cos 10^o`

Since `cos 90^o = 0` yields:

`cos100^o + cos80^o = 0`

Replacing `2cos 10^o` for `sin100^o + sin80^o` and 0 for `cos100^o + cos80^o` yields:

`(sin100^o + sin80^o) + (cos100^o + cos80^o) = 2cos 10^o`

Hence, evaluating the given summation, oerforming the conversions into products, yields `(sin100^o + sin80^o) + (cos100^o + cos80^o) = 2cos 10^o .`

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