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Basic Trigonometric Identities

If a cos^3 α ...

Question

$I f a c o s^{3} α + 3 a c o s α s i n^{2} α = m a n d a s i n^{3} α + 3 a c o s^{2} α s i n α = n, t h e n (m + n)^{\frac{2}{3}} + (m - n)^{\frac{2}{3}} i s e q u a l t o$

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Solution

The correct option is C

From the given relations, we get $m + n = a c o s^{3} α + 3 a c o s α s i n^{2} α + 3 a c o s^{2} α s i n α + a s i n^{3} α$ $= a (c o s α + s i n α)^{3}$
$S i m i l a r l y m - n = a (c o s α - s i n α)^{3} ∴ (m + n)^{\frac{2}{3}} + (m - n)^{\frac{2}{3}} = a^{\frac{2}{3}} [(c o s α + s i n α)^{2} + (c o s α - s i n α)^{2}]$

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Similar questions

$I f a c o s^{3} α + 3 a c o s α s i n^{2} α = m a n d a s i n^{3} α + 3 a c o s^{2} α s i n α = n, t h e n (m + n)^{\frac{2}{3}} + (m - n)^{\frac{2}{3}} i s e q u a l t o$

If
$a c o s^{3} α + 3 a c o s α s i n^{2} α = m a n d a s i n^{3} α + 3 a c o s^{2} α s i n α = n, t h e n (m + n)^{\frac{2}{3}} + (m - n)^{\frac{2}{3}} i s e q u a l t o$

Q. Q27

27 . I f a \cos^{3} θ + 3 a \sin^{2} θ \cos θ = m a n d a \sin^{3} θ + 3 a \sin θ \cos^{2} θ = n, p r o v e t h a t {(m + n)}^{2 / 3} + {(m - n)}^{2 / 3} = 2 a^{2 / 3}

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If a cos3 α+3a cos α sin2 α=m and a sin3 α +3a cos2 α sin α=n, then (m+n)23+(m−n)23 is equal to

The correct option is C From the given relations, we get m+n=a cos3α+3a cos α sin2α+3a cos2 α sin α+a sin3α =a(cosα+sinα)3 Similarly m−n=a(cos α−sin α)3∴ (m+n)23+(m−n)23=a23[(cos α+sin α)2+(cos α−sin α)2]

$I f a c o s^{3} α + 3 a c o s α s i n^{2} α = m a n d a s i n^{3} α + 3 a c o s^{2} α s i n α = n, t h e n (m + n)^{\frac{2}{3}} + (m - n)^{\frac{2}{3}} i s e q u a l t o$