Prove \(f \circ g=g\circ(fh)\) for \(f(x)=\sin x,\ g(x)=2x,\ h(x)=\cos x\)
📺 Video Explanation
📝 Question
Let real functions be defined as:
\[ f(x)=\sin x,\qquad g(x)=2x,\qquad h(x)=\cos x \]
Prove that:
\[ f\circ g=g\circ(fh) \]
where:
\[ (fh)(x)=f(x)\cdot h(x) \]
✅ Solution
🔹 Step 1: Find \((f\circ g)(x)\)
By definition:
\[ (f\circ g)(x)=f(g(x)) \]
Substitute \(g(x)=2x\):
\[ (f\circ g)(x)=f(2x) \]
Since:
\[ f(x)=\sin x \]
So:
\[ (f\circ g)(x)=\sin 2x \]
🔹 Step 2: Find product function \((fh)(x)\)
Since:
\[ f(x)=\sin x,\qquad h(x)=\cos x \]
Therefore:
\[ (fh)(x)=\sin x\cos x \]
🔹 Step 3: Find \((g\circ(fh))(x)\)
By definition:
\[ (g\circ(fh))(x)=g((fh)(x)) \]
Substitute:
\[ (g\circ(fh))(x)=g(\sin x\cos x) \]
Since:
\[ g(x)=2x \]
So:
\[ (g\circ(fh))(x)=2\sin x\cos x \]
🔹 Step 4: Use trigonometric identity
We know:
\[ \sin 2x=2\sin x\cos x \]
Therefore:
\[ (f\circ g)(x)=\sin 2x=2\sin x\cos x=(g\circ(fh))(x) \]
🎯 Final Answer
\[ (f\circ g)(x)=\sin 2x \]
and
\[ (g\circ(fh))(x)=2\sin x\cos x \]
Since:
\[ \sin 2x=2\sin x\cos x \]
for all real \(x\), therefore:
\[ \boxed{f\circ g=g\circ(fh)} \]
Hence proved. :contentReference[oaicite:1]{index=1}
🚀 Exam Shortcut
- First find each composition separately
- Use \((fh)(x)=f(x)\cdot h(x)\)
- Apply identity: \(\sin 2x=2\sin x\cos x\)