Question

Find the value of (i) sin 75° (ii) tan 15°

Answer 1

(i) sin 75° = sin (45° + 30°)

= sin 45° cos 30° + cos 45° sin 30°

[sin (x + y) = sin x cos y + cos x sin y]

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

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

(ii) tan 15° = tan (45°-30°)

$=\frac{tan\,45°-tan 30°}{1+tan\,45°tan 30°}$ $[tan (x-y)=\frac{tan\,x-tan\,y}{1+tan\,xtan\,y}]$

$=\frac{1-\frac{1}{\sqrt3}}{1+1(\frac{1}{√3})}=\frac{\frac{√3-1}{√3}}{\frac{√3+1}{√3}}$

$=\frac{√3-1}{√3+1}=\frac{(√3-1)^2}{(√3-1)(√3-1)}=\frac{3+1-2√3}{(√3)^2-(1)^2}$

$=\frac{4-2√3}{3-1}=2-√3$