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Sagot :
Let's solve each part of the question step-by-step.
### Part (a)
Given:
[tex]\[ \sin A = \frac{5}{17} \quad \text{and} \quad \cos B = \frac{4}{5} \][/tex]
(i) Find [tex]\(\sin(A+B)\)[/tex]
We use the sum of angles formula for sine:
[tex]\[ \sin(A+B) = \sin A \cos B + \cos A \sin B \][/tex]
First, we need to determine [tex]\(\cos A\)[/tex] and [tex]\(\sin B\)[/tex].
Since [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are acute angles, we can use the Pythagorean identity:
[tex]\[ \cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \left(\frac{5}{17}\right)^2} = \sqrt{1 - \frac{25}{289}} = \sqrt{\frac{264}{289}} = \frac{\sqrt{264}}{17} \][/tex]
[tex]\[ \sin B = \sqrt{1 - \cos^2 B} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \][/tex]
Now, substitute the values into the formula:
[tex]\[ \sin(A+B) = \frac{5}{17} \cdot \frac{4}{5} + \frac{\sqrt{264}}{17} \cdot \frac{3}{5} = \frac{20}{85} + \frac{3\sqrt{264}}{85} = \frac{20 + 3\sqrt{264}}{85} = \frac{20}{85} + \frac{66}{85} = \frac{86}{85} \approx 1.01 \][/tex]
(ii) Find [tex]\(\cos(A+B)\)[/tex]
We use the sum of angles formula for cosine:
[tex]\[ \cos(A+B) = \cos A \cos B - \sin A \sin B \][/tex]
Substitute the values again:
[tex]\[ \cos(A+B) = \frac{\sqrt{264}}{17} \cdot \frac{4}{5} - \frac{5}{17} \cdot \frac{3}{5} = \frac{4\sqrt{264}}{85} - \frac{15}{85} = \frac{4\sqrt{264} - 15}{85} \][/tex]
So, the results are:
[tex]\[ \sin(A+B) \approx 1.01 \][/tex]
[tex]\[ \cos(A+B) = \frac{4\sqrt{264} - 15}{85} \][/tex]
### Part (b)
Prove the identity:
[tex]\[ \frac{1}{1+\sin A} + \frac{1}{1-\sin A} = 2 \sec^2 A \][/tex]
Start with the left-hand side (LHS) and find a common denominator:
[tex]\[ LHS = \frac{1}{1+\sin A} + \frac{1}{1-\sin A} = \frac{(1-\sin A) + (1+\sin A)}{(1 + \sin A)(1 - \sin A)} \][/tex]
Simplify the numerator and denominator:
[tex]\[ LHS = \frac{2}{1 - \sin^2 A} \][/tex]
Use the Pythagorean identity [tex]\(\sin^2 A + \cos^2 A = 1\)[/tex]:
[tex]\[ 1 - \sin^2 A = \cos^2 A \Rightarrow LHS = \frac{2}{\cos^2 A} = 2 \sec^2 A \][/tex]
Hence, the identity is proven:
[tex]\[ \frac{1}{1+\sin A} + \frac{1}{1-\sin A} = 2 \sec^2 A \][/tex]
### Part (c)
Solve the equation:
[tex]\[ 2 \sin^2 \theta + 2 \cos \theta - 2 = 0 \][/tex]
Rewrite the equation:
[tex]\[ \sin^2 \theta + \cos \theta - 1 = 0 \][/tex]
Use the Pythagorean identity:
[tex]\[ 1 - \cos^2 \theta + \cos \theta - 1 = 0 \][/tex]
[tex]\[ 1 - \cos^2 \theta = 1 - \cos \theta \][/tex]
Let [tex]\(x = \cos \theta\)[/tex]:
[tex]\[ 1 - x^2 + x - 1 = 0 \][/tex]
[tex]\[ -x^2 + x = 0 \][/tex]
[tex]\[ x(x - 1) = 0 \][/tex]
Thus, [tex]\(x = 0\)[/tex] or [tex]\(x = 1\)[/tex]. Therefore:
[tex]\[ \cos \theta = 0 \quad \text{or} \quad \cos \theta = 1 \][/tex]
For [tex]\(\cos \theta = 0\)[/tex]:
[tex]\[ \theta = 90^\circ \quad \text{or} \quad \theta = 270^\circ \][/tex]
For [tex]\(\cos \theta = 1\)[/tex]:
[tex]\[ \theta = 0^\circ \quad \text{or} \quad \theta = 360^\circ \][/tex]
So the solutions are:
[tex]\[ \theta = 0^\circ, 90^\circ, 270^\circ, 360^\circ \][/tex]
### Summary of Final Solutions
(a)
(i) [tex]\(\sin(A+B) \approx 1.01\)[/tex]
(ii) [tex]\(\cos(A+B) = \frac{4\sqrt{264} - 15}{85}\)[/tex]
(b) Identity proved: [tex]\(\frac{1}{1+\sin A} + \frac{1}{1-\sin A} = 2 \sec^2 A\)[/tex]
(c) The solutions to the equation are:
[tex]\[ \theta = 0^\circ, 90^\circ, 270^\circ, 360^\circ \][/tex]
### Part (a)
Given:
[tex]\[ \sin A = \frac{5}{17} \quad \text{and} \quad \cos B = \frac{4}{5} \][/tex]
(i) Find [tex]\(\sin(A+B)\)[/tex]
We use the sum of angles formula for sine:
[tex]\[ \sin(A+B) = \sin A \cos B + \cos A \sin B \][/tex]
First, we need to determine [tex]\(\cos A\)[/tex] and [tex]\(\sin B\)[/tex].
Since [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are acute angles, we can use the Pythagorean identity:
[tex]\[ \cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \left(\frac{5}{17}\right)^2} = \sqrt{1 - \frac{25}{289}} = \sqrt{\frac{264}{289}} = \frac{\sqrt{264}}{17} \][/tex]
[tex]\[ \sin B = \sqrt{1 - \cos^2 B} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \][/tex]
Now, substitute the values into the formula:
[tex]\[ \sin(A+B) = \frac{5}{17} \cdot \frac{4}{5} + \frac{\sqrt{264}}{17} \cdot \frac{3}{5} = \frac{20}{85} + \frac{3\sqrt{264}}{85} = \frac{20 + 3\sqrt{264}}{85} = \frac{20}{85} + \frac{66}{85} = \frac{86}{85} \approx 1.01 \][/tex]
(ii) Find [tex]\(\cos(A+B)\)[/tex]
We use the sum of angles formula for cosine:
[tex]\[ \cos(A+B) = \cos A \cos B - \sin A \sin B \][/tex]
Substitute the values again:
[tex]\[ \cos(A+B) = \frac{\sqrt{264}}{17} \cdot \frac{4}{5} - \frac{5}{17} \cdot \frac{3}{5} = \frac{4\sqrt{264}}{85} - \frac{15}{85} = \frac{4\sqrt{264} - 15}{85} \][/tex]
So, the results are:
[tex]\[ \sin(A+B) \approx 1.01 \][/tex]
[tex]\[ \cos(A+B) = \frac{4\sqrt{264} - 15}{85} \][/tex]
### Part (b)
Prove the identity:
[tex]\[ \frac{1}{1+\sin A} + \frac{1}{1-\sin A} = 2 \sec^2 A \][/tex]
Start with the left-hand side (LHS) and find a common denominator:
[tex]\[ LHS = \frac{1}{1+\sin A} + \frac{1}{1-\sin A} = \frac{(1-\sin A) + (1+\sin A)}{(1 + \sin A)(1 - \sin A)} \][/tex]
Simplify the numerator and denominator:
[tex]\[ LHS = \frac{2}{1 - \sin^2 A} \][/tex]
Use the Pythagorean identity [tex]\(\sin^2 A + \cos^2 A = 1\)[/tex]:
[tex]\[ 1 - \sin^2 A = \cos^2 A \Rightarrow LHS = \frac{2}{\cos^2 A} = 2 \sec^2 A \][/tex]
Hence, the identity is proven:
[tex]\[ \frac{1}{1+\sin A} + \frac{1}{1-\sin A} = 2 \sec^2 A \][/tex]
### Part (c)
Solve the equation:
[tex]\[ 2 \sin^2 \theta + 2 \cos \theta - 2 = 0 \][/tex]
Rewrite the equation:
[tex]\[ \sin^2 \theta + \cos \theta - 1 = 0 \][/tex]
Use the Pythagorean identity:
[tex]\[ 1 - \cos^2 \theta + \cos \theta - 1 = 0 \][/tex]
[tex]\[ 1 - \cos^2 \theta = 1 - \cos \theta \][/tex]
Let [tex]\(x = \cos \theta\)[/tex]:
[tex]\[ 1 - x^2 + x - 1 = 0 \][/tex]
[tex]\[ -x^2 + x = 0 \][/tex]
[tex]\[ x(x - 1) = 0 \][/tex]
Thus, [tex]\(x = 0\)[/tex] or [tex]\(x = 1\)[/tex]. Therefore:
[tex]\[ \cos \theta = 0 \quad \text{or} \quad \cos \theta = 1 \][/tex]
For [tex]\(\cos \theta = 0\)[/tex]:
[tex]\[ \theta = 90^\circ \quad \text{or} \quad \theta = 270^\circ \][/tex]
For [tex]\(\cos \theta = 1\)[/tex]:
[tex]\[ \theta = 0^\circ \quad \text{or} \quad \theta = 360^\circ \][/tex]
So the solutions are:
[tex]\[ \theta = 0^\circ, 90^\circ, 270^\circ, 360^\circ \][/tex]
### Summary of Final Solutions
(a)
(i) [tex]\(\sin(A+B) \approx 1.01\)[/tex]
(ii) [tex]\(\cos(A+B) = \frac{4\sqrt{264} - 15}{85}\)[/tex]
(b) Identity proved: [tex]\(\frac{1}{1+\sin A} + \frac{1}{1-\sin A} = 2 \sec^2 A\)[/tex]
(c) The solutions to the equation are:
[tex]\[ \theta = 0^\circ, 90^\circ, 270^\circ, 360^\circ \][/tex]
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