Question

sin 3acos^3a +cos3asin^3a=3/4sin4a. prove this plesa

Answer 1

Answer:

Step-by-step explanation:

Given,

$cos^{3}A \times sin3A+sin^{3}A \times cos 3A=\frac{3}{4}sin4A$

proof,

$L.H.S=\\cos^{3}A \times sin3A+sin^{3}A \times cos 3A \:\:\:\:\:\:\:\:\:[sin3A=3sinA -4sin^{3} A]\\cos^{3}A (3sinA-4sin^{3}A )+sin^{3} A( 4 cos^{3} A - 3 cosA )\:\:\:\:\:\:\:\:\:[cos3A=4cos^{3} A-3cosA]\\3cos^{3}A sinA-4cos^{3} A sin^{3}A +4cos^{3}Asin^{3} A - 3sin^{3}A cos A \\3cos ^{3}A sinA- 3 sin^{3}A cos A\\3cosA sinA (cos^{2} A - sin^{2} A )\\3cos A sin A cos2A \:\:\:\:\:\:[ cos^{2} A - sin^{2} A=cos 2 A ]\\\frac{3}{2} \times2 cos A sin A cos2A\:\:\:\:\:\:[ sin A cos A=sin2A ]$

$\frac{3}{2\times 2} \times 2 sin2A cos2A\\\frac{3}{4} \times sin4A\:\:\:\:\:\:[ 2 sin2A cos2A=sin4A]$

hence proved

Answer 2

Answer:

$sin3Acos^{3}A + cos3Asin^{3}A = \frac{3}{4} sin4A$

Step-by-step explanation:

To prove    $sin3Acos^{3}A + cos3Asin^{3}A = \frac{3}{4} sin4A$

Trigonometric identities are based on the six trigonometric ratios. Ptolemy's identities, sum and difference formulas for sine and cosine form the basis of all trigonometric identities, proofs, relations.

Using the Trigonometric relations to prove the above equation.

Calculation:

Consider L.H.S, $sin3Acos^{3}A + cos3Asin^{3}A$

We know that $sin3A = 3sinA - 4sin^{3}A \\cos3A = 4cos^{3} A - 3cosA$.

Substituting the above formulae in the L.H.S

$(3sinA-4sin^{3} A)cos^{3}A + ( 4cos^{3}A - 3cosA)sin^{3}A\\\\3sinAcos^{3}A-4sin^{3} A cos^{3}A+4cos^{3}A sin^{3}A-3cosAsin^{3}A\\\\3sinAcos^{3}A-3cosAsin^{3}A$

Taking $3sinAcosA$ common from the two terms,

$3sinAcosA(cos^{2} A-sin^{2} A)$  .........(1)

We know that, $cos^{2} A - sin^{2}A = cos2A$ and $2sinAcosA = sin2A$

By multiplying and dividing Equation (1) by 2.

$\frac{3}{2} 2sinAcosAcos2A\\\\\frac{3}{2} sin2Acos2A\\\\\frac{3}{2*2} 2sin2Acos2A\\\\\frac{3}{4}sin4A$(R.H.S)

$sin3Acos^{3}A + cos3Asin^{3}A = \frac{3}{4} sin4A$

L.H.S = R.H.S

Hence Proved

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