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Question
Math
Posted 5 months ago

Malia tried to prove that cos(θ)=sin(90θ)\cos (\theta)=\sin \left(90^{\circ}-\theta\right) using the following diagram. Her proof is not correct.
\begin{tabular}{lll} 
Statement & Reason \\
\hline 1 & mC=90θm \angle C=90^{\circ}-\theta & The acute angles in a right triangle are complementary.
\end{tabular}
2sin(90θ)=ACBC2 \sin \left(90^{\circ}-\theta\right)=\frac{A C}{B C} \quad Definition of sine.
3cos(θ)=ACBC3 \cos (\theta)=\frac{A C}{B C} \quad Definition of cosine.
4cos(θ)=sin(90θ)4 \cos (\theta)=\sin \left(90^{\circ}-\theta\right) \quad Substitution
What is the first mistake in Malia's proof?
Choose 1 answer:
(A) The measures of B\angle B and C\angle C add up to 180180^{\circ}, not 9090^{\circ}.
(B) Malia used the wrong sides in her ratio for sin(90θ)\sin \left(90^{\circ}-\theta\right).
(C) Malia used the wrong sides in her ratio for cos(θ)\cos (\theta).
(D) Malia substituted a value that wasn't equivalent to the one she replaced.
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Answer from Sia
Posted 5 months ago
Solution by Steps
step 2
Confirm that the trigonometric identities cos(θ)=sin(90°θ)cos(\theta) = sin(90° - \theta) and sin(90°θ)=cos(θ)sin(90° - \theta) = cos(\theta) are correct
step 3
Examine Statement 1: mC=90°θm\angle C = 90° - \theta. This is correct as the angles in a right triangle are complementary
step 4
Examine Statement 2: sin(90°θ)=ACBCsin(90° - \theta) = \frac{AC}{BC}. This is the correct definition of sine for angle C
step 5
Examine Statement 3: cos(θ)=ACBCcos(\theta) = \frac{AC}{BC}. This is incorrect as the definition of cosine for angle B should be BCAC\frac{BC}{AC} or ABAC\frac{AB}{AC} depending on the triangle orientation
step 6
Identify the first mistake in Malia's proof, which is in Statement 3, where the sides used for the cosine ratio are incorrect
C
Key Concept
Trigonometric Ratios
Explanation
The cosine of an angle in a right triangle is the ratio of the adjacent side to the hypotenuse, not the opposite side to the adjacent side as used by Malia in her proof.

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