Question

a student uses the following steps to prove the sine sum identity. the proof is incorrect.
step 1 sin(x + y)
step 2 cos(\\(\frac{\pi}{2}-(x + y)\\))
step 3 cos(\\(\left(\frac{\pi}{2}-x\
ight)-y\\))
step 4 cos(\\(\frac{\pi}{2}-x\\))cos(-y)+sin(\\(\frac{\pi}{2}-x\\))sin(-y)
step 5 sin(x)cos(y)+cos(x)sin(y)
which step is incorrect?
step 2
step 3
step 4
step 5

Explanation:

Step1: Analyze Step 2

Using the co - function identity \(\sin\alpha=\cos(\frac{\pi}{2}-\alpha)\), when \(\alpha = x + y\), \(\sin(x + y)=\cos(\frac{\pi}{2}-(x + y))\). So Step 2 is correct.

Step2: Analyze Step 3

Using the associative property of addition \(\frac{\pi}{2}-(x + y)=(\frac{\pi}{2}-x)-y\). So Step 3 is correct.

Step3: Analyze Step 4

Using the cosine of a difference formula \(\cos(A - B)=\cos A\cos B+\sin A\sin B\), where \(A=\frac{\pi}{2}-x\) and \(B = y\), we have \(\cos((\frac{\pi}{2}-x)-y)=\cos(\frac{\pi}{2}-x)\cos y+\sin(\frac{\pi}{2}-x)\sin y\). Since \(\cos(-y)=\cos y\) and \(\sin(-y)=-\sin y\), the correct expansion of \(\cos((\frac{\pi}{2}-x)-y)\) should be \(\cos(\frac{\pi}{2}-x)\cos y+\sin(\frac{\pi}{2}-x)\sin y\), not \(\cos(\frac{\pi}{2}-x)\cos(-y)+\sin(\frac{\pi}{2}-x)\sin(-y)\). So Step 4 is incorrect.

Step4: Analyze Step 5 (for completeness)

If Step 4 was correct, using \(\cos(\frac{\pi}{2}-x)=\sin x\), \(\sin(\frac{\pi}{2}-x)=\cos x\), \(\cos(-y)=\cos y\) and \(\sin(-y)=-\sin y\) (but in the wrong - expanded Step 4), we would have an error. But since we already found the error in Step 4, if we assume correct trigonometric identities: \(\cos(A - B)=\cos A\cos B+\sin A\sin B\), \(A=\frac{\pi}{2}-x\), \(B = y\), \(\cos(\frac{\pi}{2}-x)=\sin x\), \(\sin(\frac{\pi}{2}-x)=\cos x\), the correct expansion \(\cos((\frac{\pi}{2}-x)-y)=\sin x\cos y+\cos x\sin y\). But the error is in Step 4.

Answer:

Step 4