prove that
cos 6A = 32cos^6A + 18cos^2A - 48cos^4A - 1
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Hey there!
Trigonometric question are a bit difficult ones O_o
But I'll try my best :p
To prove: cos 6a = 32cos^6 a - 48 cos⁴a + 18 cos² a - 1
Proof: cos6A= cos 2(3a)
= 2cos²3a -1 = 2(cos 3a)² -1
= 2 (4cos³a - 3cosa)² - 1
= 2(16 cos^6 a + 9 cos²a - 24 cos⁴a) - 1
= 32cos^6 a + 18 cos² a - 48 cos⁴ a - 1
= 32cos^6 a - 48 cos⁴ a + 18 cos² a - 1 = RHS
Hope it helped :) We got the Correct answer ^_^
Cheers!
Trigonometric question are a bit difficult ones O_o
But I'll try my best :p
To prove: cos 6a = 32cos^6 a - 48 cos⁴a + 18 cos² a - 1
Proof: cos6A= cos 2(3a)
= 2cos²3a -1 = 2(cos 3a)² -1
= 2 (4cos³a - 3cosa)² - 1
= 2(16 cos^6 a + 9 cos²a - 24 cos⁴a) - 1
= 32cos^6 a + 18 cos² a - 48 cos⁴ a - 1
= 32cos^6 a - 48 cos⁴ a + 18 cos² a - 1 = RHS
Hope it helped :) We got the Correct answer ^_^
Cheers!
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