solve in radian + degrees
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sin²π/18 ,+ sin²π/9 +sin²7π/18 + sin²4π/9 =2
sin (π/18 )= sin(π/2 -4π/9) = cos(4π/9)
{ becoz , sin(π/2 -∅) = cos∅}
sin( π/9) = sin( π/2 -7π/18) =cos(7π/18)
now,
LHS = sin²π/18 + sin²π/9 +sin²7π/18+sin²4π/9
=cos²4π/9 + cos²7π/18 + sin²7π/18 +sin²4π/9
={ sin²4π/9 +cos²4π/9} + { sin²7π/18 +cos²7π/18}
= 1 + 1 {becoz sin²x + cos²x = 1
=2 = RHS
sin (π/18 )= sin(π/2 -4π/9) = cos(4π/9)
{ becoz , sin(π/2 -∅) = cos∅}
sin( π/9) = sin( π/2 -7π/18) =cos(7π/18)
now,
LHS = sin²π/18 + sin²π/9 +sin²7π/18+sin²4π/9
=cos²4π/9 + cos²7π/18 + sin²7π/18 +sin²4π/9
={ sin²4π/9 +cos²4π/9} + { sin²7π/18 +cos²7π/18}
= 1 + 1 {becoz sin²x + cos²x = 1
=2 = RHS
abhi178:
i hope this will helpful
please can u solve that in degrees also
yeah why not
Answered by
0
Answer:
sin?u/18 ,+ sin?T1/9 +sin?717/18 + sin241/9 = 2
sin (T/18 )= sin(1/2 -41/9) = cos(41/9)
{ becoz sin(1/2-) = cos@}
sin( T1/9) = sin( T/2 -71/18) =cos(7TT/18)
now,
LHS = sin?7/18 + sin?T1/9 +sin 27T18+sin 240/9
=cos²4T/9 + cos 2717/18th + sin?71/18 +sin241/9
=( sin?4T/9 +cos²41/9} + {sin?7T/18 +cos?77/18}
= 1+1 {becoz sin?x + cos?x = 1
=2 = RHS
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