prove the identity. \n\\( \\cos \\left( x - \\frac { \\pi } { 2 } \\right) = \\sin ( x ) \\)\nthe following…

prove the identity. \n\\( \\cos \\left( x - \\frac { \\pi } { 2 } \\right) = \\sin ( x ) \\)\nthe following is a proposed proof for the identity.\n1. start with the left - hand side and first use the subtraction formula for cosine.\n2. \\( \\cos \\left( x - \\frac { \\pi } { 2 } \\right) = \\cos ( x ) \\cdot \\cos \\left( \\frac { \\pi } { 2 } \\right) + \\sin ( x ) \\cdot \\sin \\left( \\frac { \\pi } { 2 } \\right) \\)\n3. next, evaluate \\( \\cos \\left( \\frac { \\pi } { 2 } \\right) \\) and \\( \\sin \\left( \\frac { \\pi } { 2 } \\right) \\).\n4. \\( = \\cos ( x ) \\cdot 0 + \\sin ( x ) \\cdot ( - 1 ) \\)\n5. then, simplify.\n6. \\( = \\sin ( x ) \\)\n7. thus, \\( \\cos \\left( x - \\frac { \\pi } { 2 } \\right) = \\sin ( x ) \\).\nidentify the error(s) in the proposed proof. (select all that apply.)\n\\( \\square \\) line 2 expression is incorrect, using the subtraction formula for cosine results in \\( \\cos ( x ) \\cdot \\cos \\left( \\frac { \\pi } { 2 } \\right) - \\sin ( x ) \\cdot \\sin \\left( \\frac { \\pi } { 2 } \\right) \\).\n\\( \\square \\) line 3 is incorrect, we need to evaluate \\( \\cos ( \\pi ) \\) and \\( \\sin ( \\pi ) \\) instead.\n\\( \\square \\) line 4 expression is incorrect, \\( \\sin \\left( \\frac { \\pi } { 2 } \\right) = 1 \\), not \\( - 1 \\).\n\\( \\square \\) line 6 expression is incorrect, it should be \\( - \\sin ( x ) \\) instead of \\( \\sin ( x ) \\).\n\\( \\square \\) there are no errors in the proof.

prove the identity. \n\\( \\cos \\left( x - \\frac { \\pi } { 2 } \\right) = \\sin ( x ) \\)\nthe following is a proposed proof for the identity.\n1. start with the left - hand side and first use the subtraction formula for cosine.\n2. \\( \\cos \\left( x - \\frac { \\pi } { 2 } \\right) = \\cos ( x ) \\cdot \\cos \\left( \\frac { \\pi } { 2 } \\right) + \\sin ( x ) \\cdot \\sin \\left( \\frac { \\pi } { 2 } \\right) \\)\n3. next, evaluate \\( \\cos \\left( \\frac { \\pi } { 2 } \\right) \\) and \\( \\sin \\left( \\frac { \\pi } { 2 } \\right) \\).\n4. \\( = \\cos ( x ) \\cdot 0 + \\sin ( x ) \\cdot ( - 1 ) \\)\n5. then, simplify.\n6. \\( = \\sin ( x ) \\)\n7. thus, \\( \\cos \\left( x - \\frac { \\pi } { 2 } \\right) = \\sin ( x ) \\).\nidentify the error(s) in the proposed proof. (select all that apply.)\n\\( \\square \\) line 2 expression is incorrect, using the subtraction formula for cosine results in \\( \\cos ( x ) \\cdot \\cos \\left( \\frac { \\pi } { 2 } \\right) - \\sin ( x ) \\cdot \\sin \\left( \\frac { \\pi } { 2 } \\right) \\).\n\\( \\square \\) line 3 is incorrect, we need to evaluate \\( \\cos ( \\pi ) \\) and \\( \\sin ( \\pi ) \\) instead.\n\\( \\square \\) line 4 expression is incorrect, \\( \\sin \\left( \\frac { \\pi } { 2 } \\right) = 1 \\), not \\( - 1 \\).\n\\( \\square \\) line 6 expression is incorrect, it should be \\( - \\sin ( x ) \\) instead of \\( \\sin ( x ) \\).\n\\( \\square \\) there are no errors in the proof.

Answer

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 $-1$.