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tan(π cos∅) = cot( πsin∅)
we know ,
tan( π/2 - ∅) = cot∅
use this ,
cot( π/2 - πcos∅) = cot( πsin∅)
π/2 - πcos∅ = πsin∅
πsin∅ + πcos∅ = π/2
sin∅ + cos∅ = 1/2
√2{ 1/√2sin∅ + 1/√2cos∅} = 1/2
√2{ sin45°sin∅ + cos45°.cos∅} = 1/2
√2cos( ∅ - 45°) = 1/2
cos( ∅ - 45°) = 1/2√2
if we take -√2
-√2 { -1/√2 sin∅ - 1/√2cos∅ } = 1/2
-√2 { sin(-45°).sin∅ - cos( -45°).cos∅} =1/2
[ note :- cos(-∅) = cos∅ ]
-√2 cos{∅ +( -45°) } = 1/2
-√2 cos( ∅ -45°) = 1/2
cos( ∅ -45°) = -1/2√2
hence,
cos( ∅ - π/4) = ± 1/2√2
we know ,
tan( π/2 - ∅) = cot∅
use this ,
cot( π/2 - πcos∅) = cot( πsin∅)
π/2 - πcos∅ = πsin∅
πsin∅ + πcos∅ = π/2
sin∅ + cos∅ = 1/2
√2{ 1/√2sin∅ + 1/√2cos∅} = 1/2
√2{ sin45°sin∅ + cos45°.cos∅} = 1/2
√2cos( ∅ - 45°) = 1/2
cos( ∅ - 45°) = 1/2√2
if we take -√2
-√2 { -1/√2 sin∅ - 1/√2cos∅ } = 1/2
-√2 { sin(-45°).sin∅ - cos( -45°).cos∅} =1/2
[ note :- cos(-∅) = cos∅ ]
-√2 cos{∅ +( -45°) } = 1/2
-√2 cos( ∅ -45°) = 1/2
cos( ∅ -45°) = -1/2√2
hence,
cos( ∅ - π/4) = ± 1/2√2
abhi178:
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