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Question
sin6A + cos6A = 1 – 3sin2A . cos2A हे सिद्ध करा.
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Solution
डावी बाजू = sin6A + cos6A
= (sin2A)3 + (cos2A)3
= (1 – cos2A)3 + (cos2A)3 ......`[(because sin^2"A" + cos^2"A" = 1),(therefore 1 - cos^2"A" = sin^2A")]`
= 1 – 3cos2A + 3(cos2A)2 – (cos2A)3 + cos6A ......[∵ (a – b)3 = a3 – 3a2b + 3ab2 – b3]
= 1 – 3 cos2A(1 – cos2A) – cos6A + cos6A
= 1 – 3 cos2A sin2A
= उजवी बाजू
∴ sin6A + cos6A = 1 – 3sin2A . cos2A
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उकलः
Δ ABC मध्ये, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` ...(पायथागोरसचे प्रमेय)
दोन्ही बाजूला AC2 ने भागून,
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
परंतु `"AB"/"AC" = square "आणि" "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
