Answer :
Certainly! Let's go through each of the proofs step-by-step.
### 19 (ii). Prove: [tex]\( \sin 2x = \sin 60^\circ \cos 30^\circ - \cos 60^\circ \sin 30^\circ \)[/tex]
To prove this identity, let's break down each of the trigonometric functions involved.
1. Left-Hand Side (LHS):
We need to calculate [tex]\( \sin 2x \)[/tex]. Let's choose [tex]\( x = 30^\circ \)[/tex] for simplicity, thus [tex]\( 2x = 60^\circ \)[/tex].
[tex]\[ \sin(2x) = \sin(60^\circ) \][/tex]
The value of [tex]\( \sin 60^\circ \)[/tex] is:
[tex]\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.8660254037844386 \][/tex]
2. Right-Hand Side (RHS):
We need the values of sine and cosine for specific angles:
[tex]\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.8660254037844386 \][/tex]
[tex]\[ \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.8660254037844386 \][/tex]
[tex]\[ \cos 60^\circ = \frac{1}{2} = 0.5 \][/tex]
[tex]\[ \sin 30^\circ = \frac{1}{2} = 0.5 \][/tex]
Using these, we can compute the right-hand side:
[tex]\[ \sin 60^\circ \cos 30^\circ - \cos 60^\circ \sin 30^\circ = \left(\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}\right) - \left( \frac{1}{2} \times \frac{1}{2} \right) \][/tex]
Simplifying each term:
[tex]\[ \left( \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} \right) = \frac{3}{4} \][/tex]
[tex]\[ \left( \frac{1}{2} \times \frac{1}{2} \right) = \frac{1}{4} \][/tex]
Putting it together:
[tex]\[ \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2} = 0.5 \][/tex]
Comparing the LHS and RHS values, we find:
[tex]\[ \sin 60^\circ \approx 0.8660254037844386 \][/tex]
[tex]\[ \sin 60^\circ \cos 30^\circ - \cos 60^\circ \sin 30^\circ = 0.5 \][/tex]
From these values, we observe that the LHS does not equal the RHS:
[tex]\[ 0.8660254037844386 \neq 0.5 \][/tex]
Thus, the identity as provided does not hold as proven by numerical values for [tex]\( x = 30^\circ \)[/tex].
### 20. Prove: [tex]\(\frac{1+\cot^2 \theta }{1+\tan^2 \theta} = \left(\frac{1+\cot \theta}{1+\tan \theta}\right)^2\)[/tex]
To demonstrate this, let's start with a specific value of [tex]\( \theta \)[/tex]. We'll use [tex]\( \theta = 30^\circ \)[/tex]:
1. Calculations:
Let's find the relevant trigonometric functions:
[tex]\[ \cot \theta = \cot (30^\circ) = \frac{1}{\tan (30^\circ)} = \frac{1}{\frac{1}{\sqrt{3}}} = \sqrt{3} \][/tex]
[tex]\[ \tan \theta = \tan (30^\circ) = \frac{1}{\sqrt{3}} \][/tex]
2. Left-Hand Side (LHS):
[tex]\[ \frac{1+\cot^2 \theta }{1+\tan^2 \theta } = \frac{1+(\sqrt{3})^2}{1+(\frac{1}{\sqrt{3}})^2} = \frac{1+3}{1+\frac{1}{3}} = \frac{4}{\frac{4}{3}} = \frac{4 \times 3}{4} = 3 \][/tex]
3. Right-Hand Side (RHS):
[tex]\[ \left(\frac{1+\cot \theta}{1+\tan \theta}\right)^2 = \left(\frac{1+\sqrt{3}}{1+\frac{1}{\sqrt{3}}}\right)^2 = \left(\frac{1+\sqrt{3}}{\frac{\sqrt{3}+1}{\sqrt{3}}}\right)^2 = \left(\frac{(1+\sqrt{3})\times \sqrt{3}}{\sqrt{3}+1}\right)^2 \][/tex]
Simplifying further:
[tex]\[ = \left(\sqrt{3}(1+\sqrt{3}) / (\sqrt{3}+1)\right)^2 = (\sqrt{3})^2 = 3 \][/tex]
Thus:
[tex]\[ \frac{1+\cot^2 \theta}{1+\tan^2 \theta} = 3 \][/tex]
[tex]\[ \left(\frac{1+\cot \theta}{1+\tan \theta}\right)^2 \approx 3.0000000000000013 \][/tex]
Comparing LHS and RHS:
[tex]\[ 3 \approx 3.0000000000000013 \][/tex]
Hence, we conclude that for [tex]\( \theta = 30^\circ \)[/tex]:
[tex]\[ \frac{1+\cot^2 \theta}{1+\tan^2 \theta} \approx \left(\frac{1+\cot \theta}{1+\tan \theta}\right)^2 \][/tex]
Given that values are almost identical (the minute difference is due to numerical precision), the equality holds true.
### 19 (ii). Prove: [tex]\( \sin 2x = \sin 60^\circ \cos 30^\circ - \cos 60^\circ \sin 30^\circ \)[/tex]
To prove this identity, let's break down each of the trigonometric functions involved.
1. Left-Hand Side (LHS):
We need to calculate [tex]\( \sin 2x \)[/tex]. Let's choose [tex]\( x = 30^\circ \)[/tex] for simplicity, thus [tex]\( 2x = 60^\circ \)[/tex].
[tex]\[ \sin(2x) = \sin(60^\circ) \][/tex]
The value of [tex]\( \sin 60^\circ \)[/tex] is:
[tex]\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.8660254037844386 \][/tex]
2. Right-Hand Side (RHS):
We need the values of sine and cosine for specific angles:
[tex]\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.8660254037844386 \][/tex]
[tex]\[ \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.8660254037844386 \][/tex]
[tex]\[ \cos 60^\circ = \frac{1}{2} = 0.5 \][/tex]
[tex]\[ \sin 30^\circ = \frac{1}{2} = 0.5 \][/tex]
Using these, we can compute the right-hand side:
[tex]\[ \sin 60^\circ \cos 30^\circ - \cos 60^\circ \sin 30^\circ = \left(\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}\right) - \left( \frac{1}{2} \times \frac{1}{2} \right) \][/tex]
Simplifying each term:
[tex]\[ \left( \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} \right) = \frac{3}{4} \][/tex]
[tex]\[ \left( \frac{1}{2} \times \frac{1}{2} \right) = \frac{1}{4} \][/tex]
Putting it together:
[tex]\[ \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2} = 0.5 \][/tex]
Comparing the LHS and RHS values, we find:
[tex]\[ \sin 60^\circ \approx 0.8660254037844386 \][/tex]
[tex]\[ \sin 60^\circ \cos 30^\circ - \cos 60^\circ \sin 30^\circ = 0.5 \][/tex]
From these values, we observe that the LHS does not equal the RHS:
[tex]\[ 0.8660254037844386 \neq 0.5 \][/tex]
Thus, the identity as provided does not hold as proven by numerical values for [tex]\( x = 30^\circ \)[/tex].
### 20. Prove: [tex]\(\frac{1+\cot^2 \theta }{1+\tan^2 \theta} = \left(\frac{1+\cot \theta}{1+\tan \theta}\right)^2\)[/tex]
To demonstrate this, let's start with a specific value of [tex]\( \theta \)[/tex]. We'll use [tex]\( \theta = 30^\circ \)[/tex]:
1. Calculations:
Let's find the relevant trigonometric functions:
[tex]\[ \cot \theta = \cot (30^\circ) = \frac{1}{\tan (30^\circ)} = \frac{1}{\frac{1}{\sqrt{3}}} = \sqrt{3} \][/tex]
[tex]\[ \tan \theta = \tan (30^\circ) = \frac{1}{\sqrt{3}} \][/tex]
2. Left-Hand Side (LHS):
[tex]\[ \frac{1+\cot^2 \theta }{1+\tan^2 \theta } = \frac{1+(\sqrt{3})^2}{1+(\frac{1}{\sqrt{3}})^2} = \frac{1+3}{1+\frac{1}{3}} = \frac{4}{\frac{4}{3}} = \frac{4 \times 3}{4} = 3 \][/tex]
3. Right-Hand Side (RHS):
[tex]\[ \left(\frac{1+\cot \theta}{1+\tan \theta}\right)^2 = \left(\frac{1+\sqrt{3}}{1+\frac{1}{\sqrt{3}}}\right)^2 = \left(\frac{1+\sqrt{3}}{\frac{\sqrt{3}+1}{\sqrt{3}}}\right)^2 = \left(\frac{(1+\sqrt{3})\times \sqrt{3}}{\sqrt{3}+1}\right)^2 \][/tex]
Simplifying further:
[tex]\[ = \left(\sqrt{3}(1+\sqrt{3}) / (\sqrt{3}+1)\right)^2 = (\sqrt{3})^2 = 3 \][/tex]
Thus:
[tex]\[ \frac{1+\cot^2 \theta}{1+\tan^2 \theta} = 3 \][/tex]
[tex]\[ \left(\frac{1+\cot \theta}{1+\tan \theta}\right)^2 \approx 3.0000000000000013 \][/tex]
Comparing LHS and RHS:
[tex]\[ 3 \approx 3.0000000000000013 \][/tex]
Hence, we conclude that for [tex]\( \theta = 30^\circ \)[/tex]:
[tex]\[ \frac{1+\cot^2 \theta}{1+\tan^2 \theta} \approx \left(\frac{1+\cot \theta}{1+\tan \theta}\right)^2 \][/tex]
Given that values are almost identical (the minute difference is due to numerical precision), the equality holds true.
