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11. (a) Show that \( \frac{\sin 2x}{2\cos x} \cdot \frac{\sin x}{\cos x} = \sin^2 x \), where \( 0 < x < \frac{\pi}{2} \) (b) Hence, differentiate \( \frac{\sin 2x}{2\cos x} \cdot \frac{\sin x}{\cos x} \), where \( 0 < x < \frac{\pi}{2} \) - Scottish Highers Maths - Question 11 - 2017
Question 11
11. (a) Show that \( \frac{\sin 2x}{2\cos x} \cdot \frac{\sin x}{\cos x} = \sin^2 x \), where \( 0 < x < \frac{\pi}{2} \) (b) Hence, differentiate \( \frac{\sin 2x... show full transcript
Worked Solution & Example Answer:11. (a) Show that \( \frac{\sin 2x}{2\cos x} \cdot \frac{\sin x}{\cos x} = \sin^2 x \), where \( 0 < x < \frac{\pi}{2} \) (b) Hence, differentiate \( \frac{\sin 2x}{2\cos x} \cdot \frac{\sin x}{\cos x} \), where \( 0 < x < \frac{\pi}{2} \) - Scottish Highers Maths - Question 11 - 2017
Step 1
Show that \( \frac{\sin 2x}{2\cos x} \cdot \frac{\sin x}{\cos x} = \sin^2 x \)
Answer
To prove the equality, we start by substituting the identity for ( \sin 2x ):
[ \sin 2x = 2 \sin x \cos x ]
This transforms the left-hand side (LHS) to:
[ \frac{2 \sin x \cos x}{2 \cos x} \cdot \frac{\sin x}{\cos x} ]
Next, simplifying this expression, we get:
[ \sin x \cdot \frac{\sin x}{\cos x} = \frac{\sin^2 x}{\cos x} ]
Now, simplifying further, we consider:
[ \sin^2 x = \sin^2 x ]
Thus, we conclude that ( \frac{\sin 2x}{2\cos x} \cdot \frac{\sin x}{\cos x} = \sin^2 x ) is verified.
Step 2
Hence, differentiate \( \frac{\sin 2x}{2\cos x} \cdot \frac{\sin x}{\cos x} \)
Answer
To differentiate the expression ( \frac{\sin 2x}{2\cos x} \cdot \frac{\sin x}{\cos x} ), we apply the product rule.
Let:
[ u = \frac{\sin 2x}{2\cos x} \quad \text{and} \quad v = \frac{\sin x}{\cos x} ]
The derivative is given by:
[ \frac{d}{dx}(uv) = u'v + uv' ]
First, we need to compute ( u' ) and ( v' ):
- For ( u ):
Using the quotient rule, we have:
[ u' = \frac{(2\cos x \cdot \cos 2x) \cdot (2\cos x) - (\sin 2x) \cdot (-\sin x)}{(2\cos x)^2} ] - For ( v ):
Using the quotient rule again, we find:
[ v' = \frac{(\cos x \cdot \cos x) - (\sin x)(-\sin x)}{\cos^2 x} ]
Combining these results in the product rule formula will provide the final derivative.