Question

prove the identity.
\\( \cos \left( x - \frac { \pi } { 2 } \
ight) = \sin ( x ) \\)
the following is a proposed proof for the identity.

1. start with the left - hand side and first use the subtraction formula for cosine.
2. \\( \cos \left( x - \frac { \pi } { 2 } \

ight) = \cos ( x ) \cdot \cos \left( \frac { \pi } { 2 } \
ight) + \sin ( x ) \cdot \sin \left( \frac { \pi } { 2 } \
ight) \\)

1. next, evaluate \\( \cos \left( \frac { \pi } { 2 } \

ight) \\) and \\( \sin \left( \frac { \pi } { 2 } \
ight) \\).

1. \\( = \cos ( x ) \cdot 0 + \sin ( x ) \cdot ( - 1 ) \\)
2. then, simplify.
3. \\( = \sin ( x ) \\)
4. thus, \\( \cos \left( x - \frac { \pi } { 2 } \

ight) = \sin ( x ) \\).
identify the error(s) in the proposed proof. (select all that apply.)
\\( \square \\) line 2 expression is incorrect, using the subtraction formula for cosine results in \\( \cos ( x ) \cdot \cos \left( \frac { \pi } { 2 } \
ight) - \sin ( x ) \cdot \sin \left( \frac { \pi } { 2 } \
ight) \\).
\\( \square \\) line 3 is incorrect, we need to evaluate \\( \cos ( \pi ) \\) and \\( \sin ( \pi ) \\) instead.
\\( \square \\) line 4 expression is incorrect, \\( \sin \left( \frac { \pi } { 2 } \
ight) = 1 \\), not \\( - 1 \\).
\\( \square \\) line 6 expression is incorrect, it should be \\( - \sin ( x ) \\) instead of \\( \sin ( x ) \\).
\\( \square \\) there are no errors in the proof.

Explanation:

Step1: Check Line 2

The Subtraction Formula for Cosine is $\cos(A - B)=\cos A\cos B+\sin A\sin B$. Here $A = x$ and $B=\frac{\pi}{2}$, so $\cos(x-\frac{\pi}{2})=\cos x\cos\frac{\pi}{2}+\sin x\sin\frac{\pi}{2}$, line 2 is correct.

Step2: Check Line 3

We are evaluating $\cos\frac{\pi}{2}$ and $\sin\frac{\pi}{2}$, not $\cos\pi$ and $\sin\pi$. So line 3 is correct.

Step3: Check Line 4

We know that $\cos\frac{\pi}{2}=0$ and $\sin\frac{\pi}{2}=1$. So $\cos x\cdot0+\sin x\cdot1=\sin x$. The error is in line 4, where $\sin\frac{\pi}{2}$ was taken as $- 1$ instead of $1$.

Step4: Check Line 6

If line 4 was corrected ($\cos x\cdot0+\sin x\cdot1$), then line 6 would be correct. But since line 4 had an error (using $\sin\frac{\pi}{2}=-1$), if we correct line 4, line 6 is correct.

Answer:

Line 4 expression is incorrect, $\sin(\frac{\pi}{2}) = 1$, not $-1$.