Solve cos A - cos 60 - A- cos 60 - A - 14cos 3A

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Standard XII

Mathematics

Integration Using Substitution

Solve cos A...

Question

Solve $cos A . cos (60 + A) . cos (60 - A) = \frac{1}{4} cos 3 A$

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Solution

$c o s A c o s (60 - A) c o s (60 + A) = \frac{c o s 3 A}{4}$
$c o s (x - y) c o s (x + y) = c o s^{2} y - s i n^{2} x$
$c o s 3 x = 4 c o s^{3} x - 3 c o s x$
$c o s A c o s (60 - A) c o s (60 + A)$
$= c o s A (c o s^{2} (A) - s i n^{2} 60)$
$= c o s A (c o s^{2} A - \frac{3}{4})$
$= \frac{4 c o s^{3} A - 3 c o s A}{4}$
$= \frac{c o s 3 A}{4}$

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Similar questions

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A \in R

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iii)

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cos A . cos (60^{\circ} - A) cos (60^{\circ} + A) =

Q. If A =

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, find the value of

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Q. If

sec A = \frac{5}{4}

, verify that

\frac{3 sin A - 4 {sin}^{3} A}{4 {cos}^{3} A - 3 cos A} = \frac{3 tan A - {tan}^{3} A}{1 - 3 {tan}^{2} A}

If sec $A = \frac{5}{4}$ , verify that $\frac{3 s i n A - 4 s i n^{3} A}{4 c o s^{3} A - 3 c o s A} = \frac{3 t a n A - t a n^{3} A}{1 - 3 t a n^{2} A}$

Join BYJU'S Learning Program

Solve cosA.cos(60+A).cos(60−A)=14cos3A

cosAcos(60−A)cos(60+A)=cos3A4cos(x−y)cos(x+y)=cos2y−sin2xcos3x=4cos3x−3cosxcosAcos(60−A)cos(60+A)=cosA(cos2(A)−sin260)=cosA(cos2A−34)=4cos3A−3cosA4=cos3A4

Solve $cos A . cos (60 + A) . cos (60 - A) = \frac{1}{4} cos 3 A$