Question

4 sin^3 75° - 3cos 15°​

Answer 1

The value of $4sin^3$$75$°$-3cos$$15$° is  $\frac{1}{\sqrt{2} } .$

Explanation:

Given:

$4sin^3$$75$°$-3cos$$15$°

To find:

The value of $4 sin^3$$75$°-$3cos$$15$°

Solution:

$4sin^3$$75$°$-3cos$$15$°

Here, $sin$$75$° can be simplified as,

$sin$$75$°$=sin($$45$°$+30$°$)$

        $=sin45cos$$30$°$+cos45sin$

       $=\frac{1}{\sqrt{2} } .\frac{\sqrt{3} }{2}+\frac{1}{\sqrt{2} }.\frac{1}{2}$

       $=\frac{\sqrt{3} +1}{2\sqrt{2} }$

$sin^3$$75$°$=(\frac{\sqrt{3} +1}{2\sqrt{2} } )^3$

          $=\frac{(\sqrt{3}) ^3+3(\sqrt{3})^2+3\sqrt{3}+1 {} }{16\sqrt{2} }$

          $=\frac{3\sqrt{3} +9+3\sqrt{3}+1 }{16\sqrt{2} }$

          $=\frac{6\sqrt{3}+10 }{16\sqrt{2} }$

          $=\frac{2(3\sqrt{3} +5)}{16\sqrt{2} }$

          $=\frac{3\sqrt{3} +5}{8\sqrt{2} }$

$4sin^3$$75$°$=4(\frac{3\sqrt{3} +5}{8\sqrt{2} })$

$4sin^3$$75$°$=\frac{3\sqrt{3} +5}{2\sqrt{2} }$    

and $cos$$15$° can be simplified as,

$cos$$15$°$=(cos$$45$°$-cos$$30$°$)$

        $=cos45cos$$30$°$-sin45sin$

        $=\frac{1}{\sqrt{2} } .\frac{\sqrt{3} }{2} +\frac{1}{\sqrt{2} }.\frac{1}{2}$

        $=\frac{\sqrt{3}+1}{2\sqrt{2} }$

$3cos$$15$°$=3(\frac{\sqrt{3}+1}{2\sqrt{2} })$

Now substitute in the given equation,

$4sin^3$$75$°$-3cos$$15$°$=\frac{3\sqrt{3} +5}{2\sqrt{2} }-3(\frac{\sqrt{3}+1}{2\sqrt{2} })$

                         $=\frac{3\sqrt{3} +5-3\sqrt{3} -3}{2\sqrt{2} }$

                        $=\frac{2}{2\sqrt{2} }$

$4sin^3$$75$°$-3cos$$15$°$=\frac{1}{\sqrt{2} }$

So the final answer is $\frac{1}{\sqrt{2} }$.