sin 105 - sin 15

Thanks Heureka,

I have to think about that one too Chris :)

It is too late now :))

√(2) / 2

NOW GENO - I know I am a real expert at making a mountain out of an mole hill BUT how did you do it with no working?   

$$\\sin 105 - sin 15\\ =sin(180-105)-sin15\\ =sin75-sin15\\ =cos15-sin15\\ =cos(45-30)-sin(45-30)\\ =cos45cos30+sin45sin30-[sin45cos30-cos45sin30]\\ =sin45cos30+sin45sin30-sin45cos30+cos45sin30\\ =sin45sin30+sin45sin30\\ =2sin45sin30\\ =2\times\frac{1}{\sqrt2}\times\frac{1}{2}\\ =2\times\frac{1}{\sqrt2}\times\frac{1}{2}\\ =\frac{1}{\sqrt2}\\ =\frac{\sqrt2}{2}\\$$

Yeah....the way geno came up with that answer is a "sin"

Hi Chris,

I am looking forward to Geno response   LOL

Maybe there is some 'obvious' method.  But it will make me feel less stupid if you can't 'see' it either,  :))

Don't feel too stupid......I don't see the "trick," either......maybe there is some "identity" of which we are unaware....!!!

sin 105 - sin 15

$$\sin{( 105\ensurement{^{\circ}} )} - \sin{( 15\ensurement{^{\circ}} ) } \\ =\sin{( 90\ensurement{^{\circ}} + 15\ensurement{^{\circ}} )} - \sin{( 15\ensurement{^{\circ}} ) } \\ = \underbrace{\sin{( 90\ensurement{^{\circ}} ) }}_{=1}\cos{( 15\ensurement{^{\circ}} ) }+\underbrace{\cos{( 90\ensurement{^{\circ}} ) }}_{=0} \sin{( 15 \ensurement{^{\circ}}) }- \sin{( 15\ensurement{^{\circ}} ) }\\ =\cos{(\textcolor[rgb]{1,0,0}{15\ensurement{^{\circ}}} )} - \sin{( \textcolor[rgb]{1,0,0}{15 \ensurement{^{\circ}}}) } \\ \\ \boxed{ =\sqrt{2}\cos{(\textcolor[rgb]{1,0,0}{15\ensurement{^{\circ}}} +45\ensurement{^{\circ}} )}\\ =-\sqrt{2}\sin{(\textcolor[rgb]{1,0,0}{15\ensurement{^{\circ}}} -45 \ensurement{^{\circ}})} \\ =\frac{\sqrt{2}}{2} = 0.70710678119 }$$

How did you get from

cos(15) - sin(15)    to

√2cos(15 + 45)   and then to

-√2sin(15 - 45)  ????