Given \(\sin\theta+2\cos\theta=1\), show that \(|2\sin\theta-\cos\theta|=2\).
\(|2\sin\theta-\cos\theta|=2\).
Step 1: Recall a useful identity:
For any real numbers \(a, b, c, d\):
\((a\sin\theta + b\cos\theta)^2 + (c\sin\theta + d\cos\theta)^2 = (a^2+c^2)\sin^2\theta + (b^2+d^2)\cos^2\theta + 2(ab+cd)\sin\theta\cos\theta.\)
This is often used to combine expressions involving sine and cosine.
Step 2: Consider the two expressions in the question:
Step 3: Square and add them:
\[(2\sin\theta - \cos\theta)^2 + (\sin\theta + 2\cos\theta)^2.\]
Step 4: Expand each square:
\((2\sin\theta - \cos\theta)^2 = 4\sin^2\theta - 4\sin\theta\cos\theta + \cos^2\theta\)
\((\sin\theta + 2\cos\theta)^2 = \sin^2\theta + 4\sin\theta\cos\theta + 4\cos^2\theta\)
Step 5: Add them together:
\(4\sin^2\theta + \sin^2\theta = 5\sin^2\theta\)
\(\cos^2\theta + 4\cos^2\theta = 5\cos^2\theta\)
The cross terms: \(-4\sin\theta\cos\theta + 4\sin\theta\cos\theta = 0\)
So the sum is:
\(5\sin^2\theta + 5\cos^2\theta = 5(\sin^2\theta + \cos^2\theta).\)
Step 6: Recall the fundamental trigonometric identity:
\(\sin^2\theta + \cos^2\theta = 1\)
So, the sum becomes:
\(5(1) = 5\).
Step 7: We now know:
\((2\sin\theta - \cos\theta)^2 + (\sin\theta + 2\cos\theta)^2 = 5.\)
Step 8: From the question, we are given:
\(\sin\theta + 2\cos\theta = 1.\)
Therefore:
\((\sin\theta + 2\cos\theta)^2 = 1^2 = 1.\)
Step 9: Substitute this into the earlier equation:
\((2\sin\theta - \cos\theta)^2 + 1 = 5\)
Step 10: Simplify:
\((2\sin\theta - \cos\theta)^2 = 4\)
Step 11: Take square root:
\(|2\sin\theta - \cos\theta| = \sqrt{4} = 2.\)
Final Answer:
Hence, \(|2\sin\theta - \cos\theta| = 2.\)
(The absolute value is used because the expression could be +2 or -2 depending on the quadrant of \(\theta\).)