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sin25^@cos115^@=1/2(sin40^@-1)...

`sin25^@cos115^@=1/2(sin40^@-1)`

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`sin25^(@)cos115^(@)`
`=(1)/(2)(2sin 25^(@)cos115^(@))`
`=(1)/(2)[sin(25^(@)+115^(@))+sin(25^(@)-115^(@))]`
`=(1)/(2)(sin140^(@)-sin90^(@))`
`=(1)/(2)cos(180^(@)-40^(@))-1)`
`=(1)/(2)tan 40^(@)=1`
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