Find i) sin 15 ii) cos 15 using the relation
sin(A-B)= sinAcosB-cosAsinB
cos(A-B)=cosAcosB+sinAsinB
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sin(15)
= sin(45-30)
= sin 45 cos 30 - cos 45 sin 30
= 1/√2 × √3/2 - 1/√2 × 1/2
= √3/2√2 - 1/2√2
= (√3-1)/2√2
cos(15)
= cos(45-30)
= cos 45 cos 30 + sin 45 sin 30
= 1/√2 × √3/2 + 1/√2 × 1/2
= √3/2√2 + 1/2√2
= (√3+1)/2√2
= sin(45-30)
= sin 45 cos 30 - cos 45 sin 30
= 1/√2 × √3/2 - 1/√2 × 1/2
= √3/2√2 - 1/2√2
= (√3-1)/2√2
cos(15)
= cos(45-30)
= cos 45 cos 30 + sin 45 sin 30
= 1/√2 × √3/2 + 1/√2 × 1/2
= √3/2√2 + 1/2√2
= (√3+1)/2√2
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