can anybody let me know and understand why sin 75 deg. be equal this = (sq. rt of 6 + sq rt of 2)/4
Hello, rcs!
You are expected to know this identity:
. . $\displaystyle \sin(A+B) \:=\:\sin A\cos B + \sin B\cos A$
$\displaystyle \text{Show that: }\:\sin(75^o) \:=\: \frac{\sqrt{6} + \sqrt{2}}{4}$
$\displaystyle \sin(75^o) \;=\;\sin(45^o + 30^o)$
. . . . . . $\displaystyle \;=\;\sin(45^o)\cos(30^o) + \sin(30^o)(\cos(45^o)$
. . . . . . $\displaystyle =\;\left(\frac{1}{\sqrt{2}}\right)\left(\frac{ \sqrt{3}}{2}\right) + \left(\frac{1}{2}\right) \left(\frac{1}{\sqrt{2}}\right)$
. . . . . . $\displaystyle =\;\frac{\sqrt{3}+1}{2\sqrt{2}} $
. . . . . . $\displaystyle =\;\frac{\sqrt{3}+1}{2\sqrt{2}}\cdot{\color{blue} \frac{\sqrt{2}}{\sqrt{2}}} $
. . . . . . $\displaystyle =\;\frac{\sqrt{6} + \sqrt{2}}{4}$