Question

Sin75°-cos15° =????????????????

Answer 1

sin75°=cos(90°-75°)=cos15°
sin75°-cos15°=cos15°-cos15°=0

Answer 2

sin 75°=sin(45°+30°)
sin 75°=sin 45°cos 30° + cos 45°sin 30°

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

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

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

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

cos 15°=cos(60°-45°)
cos 15°=cos 60°cos 45° + sin 60°sin 45°

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

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

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

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

$\frac{ \sqrt{6} + \sqrt{2} }{4} - \frac{ \sqrt{6} + \sqrt{2} }{4} = 0$

sin 75°-cos 15°=0