Question

Answer problem #2, showing ALL work accordingly. Thank you.

Problem 2 Use the sum-difference formulas to find sin(75º), cos(75°), and tan(75)

Answer 1

we have the formulas as

$\sin(x+y)= \sin(x)\cos(y)+\cos(x)\sin(y)$

$\cos(x+y)= \cos(x)\cos(y)-\sin(x)\sin(y)$

$\tan(x+y)= \frac{\sin(x+y)}{\cos(x+y)}$

and also we have the values,

$\sin(45\degree)= \cos(45\degree)= \frac{1}{\sqrt{2}}$

$\sin(30\degree)= \frac{1}{2}$

cos(30°) V3 2

so we get the values as,

$\sin(75\degree)= \sin(45\degree+30\degree)= \sin(45\degree)\cos(30\degree)+\cos(45\degree)\sin(30\degree)$

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

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

  $= \boldsymbol{\frac{\sqrt{6}+\sqrt{2}}{4}}$

$\cos(75\degree)= \cos(45\degree+30\degree)= \cos(45\degree)\cos(30\degree)-\sin(45\degree)\sin(30\degree)$

$=( \frac{1}{\sqrt{2}}*\frac{\sqrt{3}}{2})-(\frac{1}{\sqrt{2}}*\frac{1}{2})$

  $= \boldsymbol{\frac{\sqrt{3}-1}{2\sqrt{2}}}*\frac{\sqrt{2}}{\sqrt{2}}$

$= \boldsymbol{\frac{\sqrt{6}-\sqrt{2}}{4}}$

$\tan(75\degree)= \frac{\sin(75\degree)}{\cos(75\degree)}= \frac{\sqrt{3}+1}{\sqrt{3}-1}$

  $= \frac{\sqrt{3}+1(\sqrt{3}+1)}{\sqrt{3}-1(\sqrt{3}+1)}= \frac{4+2\sqrt{3}}{3-1}$

$=\boldsymbol{ 2+\sqrt{3}}$