"Question22
Verify, 2sin45° cos45° = sin90°
Chapter6,T-Ratios of particular angles Exercise -6 ,Page number 288"
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Let's verify the following equation using the values of T-Ratios of particular angles.
We know that,
sin45 = 1/√2
cos45 = 1/√2
sin90 = 1
Now,
2 sin45 * cos45 = sin90
2 ( 1/√2 ) ( 1/√2 ) = 1
√2/√2 = 1
1 = 1
Hence, L. H. S = R. H. S.
From this, We can make a identity : 2sinx cosx = sin2x .
Thus, We proved that " 2sin45° cos45° = sin90° "
Let's verify the following equation using the values of T-Ratios of particular angles.
We know that,
sin45 = 1/√2
cos45 = 1/√2
sin90 = 1
Now,
2 sin45 * cos45 = sin90
2 ( 1/√2 ) ( 1/√2 ) = 1
√2/√2 = 1
1 = 1
Hence, L. H. S = R. H. S.
From this, We can make a identity : 2sinx cosx = sin2x .
Thus, We proved that " 2sin45° cos45° = sin90° "
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