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secondary 4 | A Maths
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Very confused on how to solve qn 7 and 12. Would appreciate detailed worked solution asap. Thanks!!
cot π/8 = cosec π/4 + cot π/4
= √2 + 1
cot π/12 = cosec π/6 + cot π/6
= 2/√3 + 1/√3
= 3/√3
= √3
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see x on left & and see x/2 on right ,
recall double angle identities
cosec x + cot x
= 1/sin 2(x/2) + 1/tan 2(x/2)
= 1/[ 2 sin (x/2) cos (x/2)] +
[ 1 - tan² (x/2) ] /[ 2 tan (x/2) ]
= [ 1/cos² (x/2) ]/[ 2 sin (x/2) cos (x/2)]/[ cos² (x/2) ] +
[ 1 - tan² (x/2) ] /[ 2 tan (x/2) ]
= [ sec² (x/2) ]/[ 2 tan (x/2) ] +
[ 1 - tan² (x/2) ] /[ 2 tan (x/2) ]
= [ 1 + tan² (x/2) ]/[ 2 tan (x/2) ] +
[ 1 - tan² (x/2) ] /[ 2 tan (x/2) ]
= 2/[ 2 tan (x/2) ]
= 1/[tan (x/2) ]
= cot (x/2) (proved)
.
from (i) ,
cosec x = cot x/2 - cot x
likewise ,
cosec 2x = cot x - cot 2x ; &
cosec 4x = cot 2x - cot 4x
hence ,
cosec x + cosec 2x + cosec 4x
= cot x/2 - cot x + cot x - cot 2x + cot 2x - cot 4x
= cot x/2 - cot 4x
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from (iii) ,
cosec 4π/15 + cosec 8π/15 + cosec 16π/15 + cosec 32π/15
= cot 4π/30 - cot 32π/15
= cot 2π/15 - cot 32π/15
= cot 2π/15 - cot (2π + 2π/15)
= cot 2π/15 - cot 2π/15
= 0
*note: 2π + 2π/15 is in 1st quad.
cot (2π + 2π/15) = cot 2π/15
.
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