Question

Verify the identity. (6 cos 0 - 2 sin 0)2 + (2 cos 0 + 6 sin 0)2 = 40 Choose the sequence of steps below that verifies the identity. B. O A. (6 cos 0 - 2 sin 0)2 + (2 cos 6 + 6 sin 0)2 = 36 cos 20 - 12 cos 0 sin 0 + 4 sin 20 + 4 cos 20 + 12 cos 0 sin 0 + 36 sin 20 = 36 (cos_0+ sin 20) +4(cos?0+ sin 20) = 36 + 4 = 40 (6 cos 0 - 2 sin 0)2 + (2 cos 0 +6 sin 0)2 = 36 cos 20 - 4 sin 20 + 4 cos 20 + 36 sin 20 = 36 (cos 20+ sin e) +4 (cos 20 + sin 20) = 36 + 4 = 40 O C. (6 cos 0 - 2 sin 0)2 + (2 cos 6 + 6 sin 0)2 = 36 cos 20-24 cos 0 sin 0 + 4 sin 20+4 cos 20 + 24 cos 0 sin 0 + 36 sin 20 = 36(cos0+ sin ²0) +4(cos 20+ sin 20) = 36 + 4 = 40

Answer 1

Given

( 6*cos( θ ) - 2*sin( θ ) )ˆ2 + ( 2*cos( θ ) + 6*sin( θ ) )ˆ2 = 40

We have to verify this equation

Check LHS

We know that --------> ( a + b )² = a² + 2*a*b + b²

( a - b )² = a² - 2*a*b + b²

Apply both the property in the LHS

( 6*cos( θ ) - 2*sin( θ ) )ˆ2 + ( 2*cos( θ ) + 6*sin( θ ) )ˆ2 = 36*cos²( θ ) - 2*6*2*cos( θ )*sin( θ ) + 4*sin²( θ ) + 4*cos²( θ ) + 2*2*6*cos( θ )*sin( θ ) + 36*sin²( θ )

=> 36*cos²( θ ) - 24*cos( θ )*sin( θ ) + 4*sin²( θ ) + 4*cos²( θ ) + 24*cos( θ )*sin( θ ) + 36*sin²( θ )

=> 36*cos²( θ ) + 4*sin²( θ ) + 4*cos²( θ ) + 36*sin²( θ )

=> 36*[ sin²( θ ) + cos²( θ ) ] + 4*[ sin²( θ ) + cos²( θ ) ]

We know that ------------> sin²( θ ) + cos²( θ ) = 1 ............................ Trigonometric Identity

=> 36*( 1 ) + 4*( 1 )

=> 36 + 4

=> 40

=> 40 = RHS

Option C is showing the correct sequence of steps