Question

Find the principal value.​

Answer 1

sin^-1[√3-1/2√2]=A

sinA=√3-1/2√2

sinA=sin15°

:sin(45°-30°)=sin45°.cos30°-cos45°.sin30°

sin(45°-30°)=(1√2)(√3/2)-(1/√2)(1/2)

sin(45°-30°)=√3-1/2√2

sin(15°)=√3-1/2√2