prove that cos20.cos40.cos80=1/8
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cos20.cos40.cos80 / 1
Multiplying Both Numerator And Denominator by 2Sin20
cos20.cos40.cos80⇒(2Sin20.cos20).cos40.cos80
---------------------------------------
2Sin20
⇒sin(2 * 20).cos40.cos80 [2SinA.CosA = Sin 2A]
-------------------------------- ↓
2sin20
⇒sin40.cos40.cos80 You Would Know This
-------------------------- If u Are In 11th or 12th
2sin20
[Multiplying both the numerator and denominator by 2]
⇒(2Sin40.Cos40).cos80
----------------------------- [2SinA.CosA = Sin 2A]
2 * 2sin20
⇒Sin(2 * 40).cos80
----------------------
4Sin20
⇒ Sin80.cos80
------------------
4Sin20
[Multiplying both the numerator and denominator by 2]
⇒2*Sin80.cos80
--------------------- [2SinA.CosA = Sin 2A]
2 * 4Sin20
⇒Sin(2 * 80)
--------------
8 Sin 20
⇒Sin 160
-----------
8 Sin20
⇒Sin (180 -20)
----------------- [Sin(180 - A) = Sin A]
8 SIn 20
⇒Sin 20 / 8 Sin 20
⇒1/8
Hope U Understood Plz Any Doubt About This Then Plz Message Me And Mark It Brainliest if this was helpful
Multiplying Both Numerator And Denominator by 2Sin20
cos20.cos40.cos80⇒(2Sin20.cos20).cos40.cos80
---------------------------------------
2Sin20
⇒sin(2 * 20).cos40.cos80 [2SinA.CosA = Sin 2A]
-------------------------------- ↓
2sin20
⇒sin40.cos40.cos80 You Would Know This
-------------------------- If u Are In 11th or 12th
2sin20
[Multiplying both the numerator and denominator by 2]
⇒(2Sin40.Cos40).cos80
----------------------------- [2SinA.CosA = Sin 2A]
2 * 2sin20
⇒Sin(2 * 40).cos80
----------------------
4Sin20
⇒ Sin80.cos80
------------------
4Sin20
[Multiplying both the numerator and denominator by 2]
⇒2*Sin80.cos80
--------------------- [2SinA.CosA = Sin 2A]
2 * 4Sin20
⇒Sin(2 * 80)
--------------
8 Sin 20
⇒Sin 160
-----------
8 Sin20
⇒Sin (180 -20)
----------------- [Sin(180 - A) = Sin A]
8 SIn 20
⇒Sin 20 / 8 Sin 20
⇒1/8
Hope U Understood Plz Any Doubt About This Then Plz Message Me And Mark It Brainliest if this was helpful
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