Show that sin105° + cos 105° = 1/√2.
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Answered by
127
We have
LHS = sin105° + cos105°
= sin(60 + 45 ) + cos ( 60 + 45 )
= ( sin 60 cos 45 + cos 60 sin 45 )
+ ( cos 60 cos 45 - sin60 din 45)
= { ( √3/2 × 1/√2 ) + ( 1/2 × 1 √2)} + { ( 1/2 × 1√2 ) - ( √3/2 × 1/√2)
= ( √3 /2√2 + 1 /2√2 + 1 2√2 + √3 /2√2 ) = 1/√2 = RHS
LHS = sin105° + cos105°
= sin(60 + 45 ) + cos ( 60 + 45 )
= ( sin 60 cos 45 + cos 60 sin 45 )
+ ( cos 60 cos 45 - sin60 din 45)
= { ( √3/2 × 1/√2 ) + ( 1/2 × 1 √2)} + { ( 1/2 × 1√2 ) - ( √3/2 × 1/√2)
= ( √3 /2√2 + 1 /2√2 + 1 2√2 + √3 /2√2 ) = 1/√2 = RHS
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