Prove that- sin A sin 60-Asin 60-A-14sin 3A

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Standard XII

Mathematics

Sine Rule

Prove that: ...

Question

Prove that:
$sin A sin (60 - A) sin (60 + A) = \frac{1}{4} sin 3 A$

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Solution

As we know that $sin (A + B) sin (A - B) = {sin}^{2} A - {sin}^{2} B$
$L H S = sin A sin (60 - A) sin (60 + A) = sin A ({sin}^{2} 60 - {sin}^{2} A) = sin A (\frac{3}{4} - {sin}^{2} A) = \frac{3 sin A - 4 {sin}^{3} A}{4} = \frac{1}{4} sin 3 A = R H S$

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Similar questions

Q. For

A \in R

, then prove that
i)

sin A sin (60 + A) sin (60 - A) = \frac{1}{4} sin 3 A

ii)

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

iii)

sin 20^{o} . sin 40^{o} . sin 60^{o} . sin 80^{o} = \frac{3}{16}

iv)

cos \frac{π}{9} . cos \frac{2 π}{9} . cos \frac{3 π}{9} . cos \frac{4 π}{9} = \frac{1}{16}

Provie that $s i n A s i n (60^{\circ} - A) s i n (60^{\circ} + A) = \frac{4}{4} s i n 3 A .$

Hene , deduce that $s i n 20^{\circ} s i n 40^{\circ} s i n 60^{\circ} s o m 80^{\circ} = \frac{3}{16} .$

Prove tahat $c o t A + c o t (60^{\circ} + A) - c o t (60^{\circ} - A) = 3 c o t 3 A .$

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 $\frac{3 s i n A - 4 s i n^{3} A}{4 c o s^{3} A - 3 c o s A} = \frac{3 t a n A - t a n^{3} A}{1 - 3 t a n^{2} A}$

If 2 sin A – 1 = 0, then: 3 Sin A – 4 Sin³A = ?

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Prove that:sinAsin(60−A)sin(60+A)=14sin3A

As we know that sin(A+B)sin(A−B)=sin2A−sin2BLHS=sinAsin(60−A)sin(60+A)=sinA(sin260−sin2A)=sinA(34−sin2A)=3sinA−4sin3A4=14sin3A=RHS

Prove that:
$sin A sin (60 - A) sin (60 + A) = \frac{1}{4} sin 3 A$

As we know that $sin (A + B) sin (A - B) = {sin}^{2} A - {sin}^{2} B$
$L H S = sin A sin (60 - A) sin (60 + A) = sin A ({sin}^{2} 60 - {sin}^{2} A) = sin A (\frac{3}{4} - {sin}^{2} A) = \frac{3 sin A - 4 {sin}^{3} A}{4} = \frac{1}{4} sin 3 A = R H S$