Question

Question 13 Prove that cos2 2x – cos2 6x = sin 4x sin 8x

Class X1 - Maths -Trigonometric Functions Page 73

Answer 1

LHS =cos²(2x) - cos²(6x)
Use the formula,
a² - b² = (a - b)(a + b)
= (cos2x - cos6x)(cos2x + cos6x)
Use the formula
cosC + cosD = 2cos(C+D)/2.cos(C-D)/2
cosC - cosD = 2sin(C+D)/2.sin(D-C)/2

= {2sin(2x + 6x)/2.sin(6x-2x)/2}{2cos(2x+6x)/2.cos(6x -2x)/2}
={2sin2x.sin4x}{2cos4x.cos2x}
={2sin2x.cos2x}{2sin4x.cos4x}

Now,
Use the formula,
sin2A = 2sinA.cosA

= sin2(2x).sin2(4x)
= sin4x.sin8x = RHS

Answer 2

cos²2x-cos²6x

=(cos2x+cos6x)(cos2x-sin2x)

=2cos(2x+6x/2)cos(6x-2x/2)*2sin(2x+6x/2)sin(6x-2x/2)

=2cos4xcos2x*2sin4xcos2x

=2cos4x*sin4x(2cos2x*sin2x)

=sin2(4)x*sin2(2)x

=sin8x*sin4x

=sin4x*sin8x

Formula :-
a+b ( a-b) = a^2-b^2

cos(c+d) = 2 cos ( c + d /2) cos (c-d/2)

cos(c-d) = 2 sin( c + d /2) sin (d-c/2)

sin2∅ = 2sin∅ cos∅