Question

noah tried to prove that \\(\cos(\theta) = \sin(\theta)\\) using the following diagram. his proof is not correct.

Explanation:

To determine why Noah's proof that \(\cos(\theta)=\sin(\theta)\) is incorrect, we analyze the right triangle and trigonometric ratios:

Step 1: Recall Trigonometric Ratios

In a right triangle, for an acute angle \(\theta\):

• \(\cos(\theta)=\frac{\text{adjacent side to } \theta}{\text{hypotenuse}}\)
• \(\sin(\theta)=\frac{\text{opposite side to } \theta}{\text{hypotenuse}}\)

Step 2: Identify Sides for \(\theta\) at \(C\)

In \(\triangle ABC\) (right-angled at \(A\)):

• Adjacent side to \(\theta\) (at \(C\)): \(AC\)
• Opposite side to \(\theta\) (at \(C\)): \(AB\)
• Hypotenuse: \(BC\)

So, \(\cos(\theta)=\frac{AC}{BC}\) and \(\sin(\theta)=\frac{AB}{BC}\). For \(\cos(\theta)=\sin(\theta)\), we need \(AC = AB\), meaning the triangle must be isosceles (legs \(AC\) and \(AB\) equal). However, Noah’s proof assumes \(\cos(\theta)=\sin(\theta)\) generally, but this equality only holds for \(\theta = 45^\circ\) (in a right isosceles triangle). His error is assuming the ratio holds for any right triangle, not just isosceles ones.

Answer:

Noah’s proof is incorrect because \(\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{AC}{BC}\) and \(\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{AB}{BC}\). These are equal only if \(AC = AB\) (isosceles right triangle, \(\theta = 45^\circ\)), not for all right triangles. He incorrectly generalizes the equality to all \(\theta\) in right triangles.