prove that sin six thita + cos six thita = 1-3 sin square thita.cos square thita.
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to prove
sin6ө + cos6ө =1-3sin2өcos2ө
(sin²Ф)³ + (cos ²Ф)³
∴ a³+b³= (a+b)³-3ab(a+b)
(sin²Ф + cos²Ф)³ - 3(sin²Ф * cos²Ф)(sin²Ф + cos²Ф)
acc. to sin²Ф + cos²Ф = 1
(1)³-3(sin²Ф * cos²Ф)(1)
1-3(sin²Ф * cos²Ф)
sin6ө + cos6ө =1-3sin2өcos2ө
(sin²Ф)³ + (cos ²Ф)³
∴ a³+b³= (a+b)³-3ab(a+b)
(sin²Ф + cos²Ф)³ - 3(sin²Ф * cos²Ф)(sin²Ф + cos²Ф)
acc. to sin²Ф + cos²Ф = 1
(1)³-3(sin²Ф * cos²Ф)(1)
1-3(sin²Ф * cos²Ф)
Adarsh9450668057:
a lot of thankyou.
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