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)\(\frac{\sin \Theta\left ( \sin \Theta -\cos \Theta \right )+\cos \Theta \left ( \sin \Theta +\cos \Theta \right ) }{\left ( \sin \Theta +\cos \Theta \right )\left ( \sin \Theta -\cos \Theta \right )}\)
\(\frac{\sin^{2}\Theta -\sin \Theta \cos \Theta +\sin \Theta \cos \Theta +\cos ^{2}\Theta }{\sin ^{2}\Theta -\cos ^{2}\Theta }\)
\(\frac{1}{\sin ^{2}\Theta -\cos ^{2}\Theta }\)
(ii)\(\sin ^{2}\Theta -\left ( 1-\sin ^{2} \Theta \right )=\frac{1}{3}\) or \(1-\cos ^{2}\Theta -\cos ^{2}\Theta =\frac{1}{3}\)
or \(3\left ( \sin ^{2} \Theta -\cos ^{2}\Theta \right )=\cos ^{2}\Theta +\sin ^{2}\Theta \)
\(\sin \Theta =\pm \sqrt{\frac{2}{3}}\) or \(\cos \Theta =\pm \sqrt{\frac{1}{3}}\)
\(\tan \Theta =\pm \sqrt{2}\)
\(\Theta =54.7^{\circ},125.3^{\circ},234.7^{\circ},305.3^{\circ}\)
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)\(\cos ^{4}x=\left ( 1-\sin ^{2}x \right )^{2}=1-2\sin ^{2}x+\sin ^{4}x\)
(ii)\(8\sin ^{4}x+1-2\sin ^{2}x+\sin ^{4}x=2\left ( 1-\sin ^{2}x \right )\)
\(9\sin ^{4}x=1\)
\(x=35.3^{\circ}\) (or any correct solution)
Any correct second solution from \(144.7^{\circ},215.3^{\circ},324.7^{\circ}\)
The remaining 2 solutions
Question
Solve the equation \(\sin ^{-1}\left ( 4x^{4}+x^{2} \right )=\frac{1}{6}\pi \).
Answer/Explanation
\(4x^{2}+x^{2}=\frac{1}{2}\)
Solve as quadratic in \(x^{2}\)
\(x^{2}=\frac{1}{4}\)
\(x=\pm \frac{1}{2}\)
Question
Find the value of x satisfying the equation \(\sin ^{-1}\left ( x-1 \right )=\tan ^{-1}3\).
Answer/Explanation
\(\tan ^{-1}(3)=1.249\) or \(71.565^{\circ}\)
\(\sin 1.25\) \( \sin 71.6\) or 0.949
x=1.95 accept
\(1+\frac{3}{\sqrt{10}}\)
Question.
(a) Solve the equation \(sin^{-1}(3x) = -1\), giving the solution in an exact form.
(b) Solve, by factorising, the equation \(2 cos\Theta sin\Theta – 2 cos\Theta – sin\Theta +1 = 0\) for \(0 \leq \Theta \leq \Pi\).
Answer/Explanation
Ans:(a)\( (3x)=-\frac{\sqrt{3}}{2}\rightarrow x=\frac{-\sqrt{3}}{6} \)
(b)\( (2cos\Theta -1)(sin\Theta -1)=0\)
\(cos=\frac{1}{2} or sin\Theta =1\)
\(\Theta =\frac{\Pi }{3}\) or \(\frac{\Pi }{2}\)