Trigonometry Question

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For the following proof, determine the missing statements and reasons in the questions below.

Let circle O be the unit circle, whose center is the origin. Let

$theta$ and

$alpha$ be two angles in standard position, such that

$0 < alpha < theta < 2pi$ . Let point

$A$ be the intersection of the terminal arm of angle

$theta$ and circle O, with coordinates

$(x_A, y_A)$ . Let point

$B$ be the intersection of the terminal arm of angle

$alpha$ and circle O, with coordinates

$(x_B, y_B)$ . Prove that

$cos(theta-alpha) = cos(theta)cos(alpha) + sin(theta)sin(alpha)$ .

$\ \ \ \ \ \ \ \ \ \ \ \ " Statement " \ \ \ \ \ \ \ \ \ \ \ \$	$" Reason "$
$1. m ang AOB + alpha = theta \ \ \ \ \ \ \ \ \ \ \$	$1. "Angle Addition Postulate"$
$2. m ang AOB = theta - alpha$	$2. "Subtraction Property of Equality"$
$3. AB^2 = OA^2 + OB^2 - 2OA \ OBcos(m ang AOB)$	$3. ""$
$4. AB^2 = OA^2 + OB^2 - 2OA \ OBcos(theta-alpha)$	$4. "Substitution Property of Equality"$
$5.$	$5. "Definition of radius"$
$6. OA = OB = 1$	$6. "Given (circle O is the unit circle)"$
$7. AB^2 = 1+1-2(1)(1)Cos(theta-alpha)$	$7. ""$
$8. AB^2 = 2-2cos(theta-alpha)$	$8. "Simplification (add, multiply)"$
$9. (x_A, y_A) = (cos(theta), sin(theta))$ , $\ \ \ (x_B, y_B) = (cos(alpha), sin(alpha))$	$9. "For any angle " gamma " in standard position,"$ $\ \ \ "the coordinates of the intersection"$ $\ \ \ "of its terminal arm and the unit circle"$ $\ \ \ "are " (cos(gamma), sin(gamma))$
$10. AB = sqrt((x_B - x_A)^2 + (y_B - y_A)^2)$	$10. ""$
$11. AB^2 = (x_B - x_A)^2 + (y_B - y_A)^2$	$11. "Multiplication Property of Equality"$
$12. AB^2 = (cos(alpha)-cos(theta))^2 + (sin(alpha)-sin(theta))^2$	$12. "Substitution Property of Equality"$
$13.$	$13. "Distributive Property of Equality"$
$14. AB^2 = cos^2(theta) + sin^2(theta) + cos^2(alpha) + sin^2(alpha)$ $\ \ \ - 2cos(theta)cos(alpha) - 2sin(theta)sin(alpha)$	$14. "Commutative Property of Addition,"$ $\ \ \ "Commutative Property of Multiplication"$
$15. AB^2 = 1 + 1 - 2cos(theta)cos(alpha) - 2sin(theta)sin(alpha)$	$15. ""$
$16. AB^2 = 1 + 1 - 2[ \ cos(theta)cos(alpha) + sin(theta)sin(alpha) \ ]$	$16. "Distributive Property of Equality"$
$17. AB^2 = 2 - 2[ \ cos(theta)cos(alpha) + sin(theta)sin(alpha) \ ]$	$17. "Simplification (add)"$
$18. 2 - 2cos(theta-alpha) = 2 - 2[ \ cos(theta)cos(alpha) + sin(theta)sin(alpha) \ ]$	$18. ""$
$19. - 2cos(theta-alpha) = - 2[ \ cos(theta)cos(alpha) + sin(theta)sin(alpha) \ ]$	$19. "Subtraction Property of Equality"$
$20. cos(theta-alpha) = cos(theta)cos(alpha) + sin(theta)sin(alpha)$	$20. "Division Property of Equality"$

Note: If

$theta - alpha= m ang AOB > pi$ , then the equation in step 3 will be

$AB^2 = OA^2 + OB^2 - 2OA \ OB cos(2pi - m ang AOB)$ . However, since

$cos(2pi - x) = cos(x), forall x in RR$ , then

$cos(2pi - m ang AOB) = cos(m ang AOB)$ , and the proof remains the same.
Also, if

$theta - alpha = m ang AOB = pi$ , this proof is not valid. However, the equation to be proved simply becomes, on the LHS,

$-1$ , and on the RHS

$cos(theta)cos(alpha)+sin(theta)sin(alpha) =$

$cos(pi + alpha)cos(alpha) + sin(pi + alpha)sin(alpha) =$

$-cos^2(alpha) - sin^2(alpha) = -1$ , proving the equation is true for this special case as well.

Grade 11 Trigonometry CCSS: HSF-TF.C.9

What is the missing statement in step 13?

$AB^2 = cos^2(alpha)sin^2(alpha) + cos^2(theta) sin^2(theta) - 2cos(alpha)sin(alpha)cos(theta)sin(theta)$
$AB^2 = cos^2(alpha) - 2cos(alpha)sin(alpha) + sin^2(alpha) + cos^2(theta) - 2cos(theta)sin(theta) + sin^2(theta)$
$AB^2 = cos^2(alpha) - 2cos(alpha)cos(theta) + cos^2(theta) + sin^2(alpha) - 2sin(alpha)sin(theta) + sin^2(theta)$
$AB^2 = cos^2(alpha) + cos^2(theta) + sin^2(alpha) + sin^2(theta)$

Question Info

Trigonometry Question

Grade 11 Trigonometry CCSS: HSF-TF.C.9