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by **jazz98** on Thu Apr 15, 2010 6:49 am

$sin75^o=sin(45^o+30^o)$
$sin(45^o+30^o)=sin45^o cos30^o+cos45^o sin30^o$
$=\left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)+\left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right)$
$=\left(\sqrt{3}+1\right)\left(\frac{\sqrt{2}}{4}\right)$

someone please let me know, why $=sin45^o cos30^o+cos45^o sin30^o$
become $=\left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)+\left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right)$

where to get the formulae

**Sponsor**

by **mathnoob** on Thu Apr 15, 2010 10:59 am

they should have givn you a list of "reference" angles for 0, pi/6, pi/4, pi/3, pi/2
you can see it here: http://www.purplemath.com/modules/trig.htm
they just plugged the numbers in like sin pi/4 = sqrt2 / 2 you just memorize it

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Basic trigonometric identities

Basic trigonometric identities

Re: Basic trigonometric identities