Question

If sin 5A = cos 4 A, then 2 sin 3 A - √3cos3A​

Answer 1

32sinA/2.sin5A/2

32sinA/2.sin5A/2 use formula ,

32sinA/2.sin5A/2 use formula ,2sinA.sinB = cos(A -B) -cos(A+B)

32sinA/2.sin5A/2 use formula ,2sinA.sinB = cos(A -B) -cos(A+B) cos2A = 2cos²A -1

32sinA/2.sin5A/2 use formula ,2sinA.sinB = cos(A -B) -cos(A+B) cos2A = 2cos²A -1cos3A = 4cos³A -3cosA

32sinA/2.sin5A/2 use formula ,2sinA.sinB = cos(A -B) -cos(A+B) cos2A = 2cos²A -1cos3A = 4cos³A -3cosA now,

32sinA/2.sin5A/2 use formula ,2sinA.sinB = cos(A -B) -cos(A+B) cos2A = 2cos²A -1cos3A = 4cos³A -3cosA now, 32sinA/2sin5A/2 = 16cos2A -16cos3A

32sinA/2.sin5A/2 use formula ,2sinA.sinB = cos(A -B) -cos(A+B) cos2A = 2cos²A -1cos3A = 4cos³A -3cosA now, 32sinA/2sin5A/2 = 16cos2A -16cos3A16{ 2cos²A -1 -4cos³A+3cosA}

32sinA/2.sin5A/2 use formula ,2sinA.sinB = cos(A -B) -cos(A+B) cos2A = 2cos²A -1cos3A = 4cos³A -3cosA now, 32sinA/2sin5A/2 = 16cos2A -16cos3A16{ 2cos²A -1 -4cos³A+3cosA} =16{ 2(3/4)²-1-4(3/4)³+3(3/4)}

32sinA/2.sin5A/2 use formula ,2sinA.sinB = cos(A -B) -cos(A+B) cos2A = 2cos²A -1cos3A = 4cos³A -3cosA now, 32sinA/2sin5A/2 = 16cos2A -16cos3A16{ 2cos²A -1 -4cos³A+3cosA} =16{ 2(3/4)²-1-4(3/4)³+3(3/4)} =16{ 18/16 -1 -27/16 +36/16}

32sinA/2.sin5A/2 use formula ,2sinA.sinB = cos(A -B) -cos(A+B) cos2A = 2cos²A -1cos3A = 4cos³A -3cosA now, 32sinA/2sin5A/2 = 16cos2A -16cos3A16{ 2cos²A -1 -4cos³A+3cosA} =16{ 2(3/4)²-1-4(3/4)³+3(3/4)} =16{ 18/16 -1 -27/16 +36/16}=( 18 -16 -27 +36}

32sinA/2.sin5A/2 use formula ,2sinA.sinB = cos(A -B) -cos(A+B) cos2A = 2cos²A -1cos3A = 4cos³A -3cosA now, 32sinA/2sin5A/2 = 16cos2A -16cos3A16{ 2cos²A -1 -4cos³A+3cosA} =16{ 2(3/4)²-1-4(3/4)³+3(3/4)} =16{ 18/16 -1 -27/16 +36/16}=( 18 -16 -27 +36}=11

by BRAINLY

Answer 2

Answer:

32sinA/2.sin5A/2

use formula ,

2sinA.sinB = cos(A -B) -cos(A+B)

cos2A = 2cos²A -1

cos3A = 4cos³A -3cosA

now,

32sinA/2sin5A/2 = 16cos2A -16cos3A

16{ 2cos²A -1 -4cos³A+3cosA}

=16{ 2(3/4)²-1-4(3/4)³+3(3/4)}

=16{ 18/16 -1 -27/16 +36/16}

=( 18 -16 -27 +36}

=11

It might help you

it was VERY hard for me as I have solved this type of question after a long time when I was studying .

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