Question
(i)Show that \( \frac{\sin \Theta }{\sin \Theta +\cos \Theta }+\frac{\cos \Theta }{\sin \Theta -\cos \Theta }=\frac{1}{\sin ^{2}\Theta -\cos ^{2}\Theta }\).
(ii)Hence solve the equation \(\frac{\sin \Theta }{\sin \Theta +\cos \Theta }+\frac{\cos \Theta }{\sin \Theta -\cos \Theta }=3\) for \(0^{\circ}\leq \Theta \leq 360^{\circ}\)
▶️Answer/Explanation
(i)Solving LHS
\( \frac{\sin \Theta}{\sin \Theta + \cos \Theta} + \frac{\cos \Theta}{\sin \Theta – \cos \Theta} \)
\( = \frac{\sin \Theta(\sin \Theta – \cos \Theta) + \cos \Theta(\sin \Theta + \cos \Theta)}{(\sin \Theta + \cos \Theta)(\sin \Theta – \cos \Theta)} \)
\( = \frac{\sin^2 \Theta – \sin \Theta \cos \Theta + \sin \Theta \cos \Theta + \cos^2 \Theta}{\sin^2 \Theta – \cos^2 \Theta} \)
\( = \frac{\sin^2 \Theta + \cos^2 \Theta}{\sin^2 \Theta – \cos^2 \Theta} \)
\( = \frac{1}{\sin^2 \Theta – \cos^2 \Theta} \)
(ii) \( \frac{1}{\sin^2 \Theta – \cos^2 \Theta} = 3 \)
\( \sin^2 \Theta – \cos^2 \Theta = \frac{1}{3} \)
\( \sin^2 \Theta – (1 – \sin^2 \Theta) = \frac{1}{3} \] (\(\cos^2 \Theta = 1 – \sin^2 \Theta\))
\( 2\sin^2 \Theta – 1 = \frac{1}{3} \)
\( 2\sin^2 \Theta = \frac{4}{3} \)
\( \sin^2 \Theta = \frac{2}{3} \)
\( \sin \Theta = \pm\sqrt{\frac{2}{3}} \)
$\Theta=54.7^{\circ}, 125.3^{\circ}, 234.7^{\circ}, 305.3^{\circ}$
Question
(i) Show that \(\cos ^{4}x=1-2\sin ^{2}x+\sin ^{4}x.\)
(ii)Hence,or otherwise,solve the equation \(8\sin ^{4}x+\cos ^{4}x=2\cos ^{2}x \) for \(0^{\circ}\leq x\leq 360^{\circ}\)
▶️Answer/Explanation
(i) \(\cos^4 x = (\cos^2 x)^2\)
\(= (1 – \sin^2 x)^2\) \
\(= 1 – 2\sin^2 x + \sin^4 x\)
Hence,proved.
(ii) \(8 \sin^4 x + (1 – 2 \sin^2 x + \sin^4 x) = 2 \cos^2 x\)
\(8 \sin^4 x + 1 – 2 \sin^2 x + \sin^4 x = 2(1 – \sin^2 x)\) (again using \(\cos^2 x = 1 – \sin^2 x\))
\(9 \sin^4 x – 2 \sin^2 x + 1 = 2 – 2 \sin^2 x\)
\(9 \sin^4 x – 2 \sin^2 x + 1 – 2 + 2 \sin^2 x = 0\)
\(9 \sin^4 x – 1 = 0\)
\(\sin^4 x = \frac{1}{9}\)
\(\sin^2 x = \pm\sqrt{\frac{1}{9}}\)
\(\sin^2 x = \pm\frac{1}{3}\)
As, \(\sin^2 x\) cannot be negative, we will only consider the positive solution:
\(\sin^2 x = \frac{1}{3}\)
\(\sin x = \pm\sqrt{\frac{1}{3}}\)
\(\sin x = \pm\frac{1}{\sqrt{3}}\)
For \(\sin x = \frac{1}{\sqrt{3}}\):
\(x = \sin^{-1}\left(\frac{1}{\sqrt{3}}\right)\) (First Quadrant)
\(x = 180^\circ – \sin^{-1}\left(\frac{1}{\sqrt{3}}\right)\) (Second Quadrant)
For \(\sin x = -\frac{1}{\sqrt{3}}\):
\(x = 180^\circ + \sin^{-1}\left(\frac{1}{\sqrt{3}}\right)\) (Third Quadrant)
\(x = 360^\circ – \sin^{-1}\left(\frac{1}{\sqrt{3}}\right)\) (Fourth Quadrant)
\Theta will have values $35.3^{\circ}$,$144.7^{\circ}, 215.3^{\circ}, 324.7^{\circ}$
Question
Solve the equation \(\sin ^{-1}\left ( 4x^{4}+x^{2} \right )=\frac{1}{6}\pi \).
▶️Answer/Explanation
\( 4x^4 + x^2 = \sin\left(\frac{\pi}{6}\right) \)
\( 4x^4 + x^2 = 0.5 \)
\( 4y^2 + y – 0.5 = 0 \)
\( y = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \)
where \(a = 4\), \(b = 1\), and \(c = -0.5\).
\( y = \frac{-1 \pm \sqrt{1 – 4 \cdot 4 \cdot -0.5}}{2 \cdot 4} \)
\( y = \frac{-1 \pm \sqrt{1 + 8}}{8} \)
\( y = \frac{-1 \pm \sqrt{9}}{8} \)
\( y = \frac{-1 \pm 3}{8} \)
\( y = \frac{-1 + 3}{8} = \frac{2}{8} = \frac{1}{4} \)
\( y = \frac{-1 – 3}{8} = \frac{-4}{8} = -\frac{1}{2} \)
However, \(y = x^2\) can not be negative.Thus, \(x^2 = \frac{1}{4}\),
\( x = \pm\sqrt{\frac{1}{4}} \)
\( x = \pm\frac{1}{2} \)
So, the solutions to the original equation are \(x = \frac{1}{2}\) and \(x = -\frac{1}{2}\).