Question
(i)Prove the identity \(\frac{\sin \Theta }{1-\cos \Theta }-\frac{1}{\sin \Theta }=\frac{1}{\tan \Theta }\).
(ii) Hence solve the equation \(\frac{\sin \Theta }{1-\cos \Theta }-\frac{1}{\sin \Theta }=4\tan \Theta \) for \(0^{\circ}< \Theta < 180^{\circ}\).
Answer/Explanation
(i)\(LHS=\frac{\sin ^{2}\Theta -(1-\cos \Theta )}{\left ( 1-\cos \Theta \right )\sin \Theta }\)\
\(=\frac{1-\cos ^{2}\Theta-1+\cos \Theta }{\left ( 1-\cos \Theta \right )\sin \Theta }\)
\(=\frac{\cos \Theta \left ( 1-\cos \Theta \right ) }{\left ( 1-\cos \Theta \right )\sin \Theta }\)
\(=\frac{1}{\tan \Theta }\)
(ii)\(\tan \Theta =\pm \frac{1}{2}\)
\(26.6^{\circ},153.4^{\circ}\)
Question
(a)The diagram shows part of the graph of \(y=a+b\sin x\).Find the values of constant a and b.
(b)(i)Show that the equation \(\left ( \sin \Theta +2\cos \Theta \right )\left ( 1+\sin \Theta -\cos \Theta \right )=\sin \Theta \left ( 1+\cos \Theta \right )\)
may be expressed as \(3\cos ^{2}\Theta -2\cos \Theta -1=0\).
(c)Hence solve the equation \(\left ( \sin \Theta +2\cos \Theta \right )\left ( 1+\sin \Theta-\cos \Theta \right )=\sin \Theta \left ( 1+\cos \Theta \right )\) for \(-180^{\circ}\leq \Theta \leq 180^{\circ}\)
Answer/Explanation
(i)a=-2,b=3
(b)(ii)\(s+s^{2}-sc+2c+2sc=s+sc\rightarrow s^{2}-2c^{2}+2c=0\)
\(1-\cos ^{2}\Theta -2\cos ^{2}\Theta +2\cos \Theta =0\)
\(3\cos ^{2}\Theta -2\cos \Theta -1=0\)
(ii)\(\cos \Theta = 1\)or \( -\frac{1}{3}\)
\(\Theta =0^{\circ}\) or \(109.5^{\circ}\) or \(-109.5^{\circ}\)
Question
(i) Show that the equation \( \frac{4\cos \Theta }{\tan \Theta }+15=0\) can be expressed as \(4\sin ^{2}\Theta -15\sin \Theta -4=0\).
(ii) Hence solve the equation \(\frac{4\cos \Theta }{\tan \Theta }+15=0\) for \(0^{\circ}\leq \Theta \leq 360^{\circ}\).
Answer/Explanation
(i)\(4\cos ^{2}\Theta +15\sin \Theta =0\)
\(4(1-s^{2})+15s=0\rightarrow 4\sin ^{2}\Theta -15\sin \Theta -4=0\)
(ii)\(\sin \Theta =-\frac{1}{4}\)
\(\Theta \)=194.5 or 345.5
Question
Solve the equation \(3 sin^2\Theta = 4 cos \Theta -1\) for 0° < 0 ≤ 360°.
Answer/Explanation
\(3\sin ^{2}\Theta =4\cos \Theta -1\)
Uses \(s^{2}+c^{2}=1\)
\(\rightarrow 3c^{2}+4c-4\left ( =0 \right )\)
\(\left ( \rightarrow c=\frac{2}{3} or -2\right )\)
\(\rightarrow \Theta =48.2^{\circ}\) or \(311.8^{\circ}\)
0.841,5.44 rads,A1 only
\(\left ( 0.268\pi ,1.73\pi \right )\)
Question
(i) Prove the identity \(\left ( \frac{1}{cos}-tanx \right )^{2}\equiv \frac{1-sinx}{1+sinx}\)
(ii) Hence solve the equation \(\left ( \frac{1}{cosx}-tan2x \right )^{2}=\frac{1}{3}\) for0 ≤ x ≤ 0.
Answer/Explanation
(i) LHS=\(\left ( \frac{1}{c} -\frac{s}{c}\right )^{2}=\frac{(1-s)(1-s)}{c^{2}}\)
=\(\frac{(1-s)(1-s)}{1-s^{2}}\)
=\(\frac{(1-s)(1-s)}{(1-s)(1+s)}\)
\(\frac{1-sinx}{1+sinx}\)
(ii) Uses part (i) to obtain \(\frac{1-sin2x}{1+sin2x}=\frac{1}{3}\rightarrow sin2x=\frac{1}{2}
x=\frac{\Pi }{12}
x=\frac{5\Pi }{12}\)
Question
(a) Show that \(\frac{\sin \theta }{1-\sin \theta }-\frac{\sin \theta }{1+\sin \theta }=2\tan ^{2}\theta \) [3]
(b) Hence solve the equation \(\frac{\sin \theta }{1-\sin \theta }-\frac{\sin \theta }{1+\sin \theta }=8, for\ 0^{o}< \theta < 180^{o}\) [3]
Answer/Explanation
Ans
7 (a) \(\left ( \frac{\sin \theta }{1-\sin \theta } -\frac{\sin \theta }{1+\sin \theta }\right )=\frac{\sin \theta (1+\sin \theta )-\sin \theta (1-\sin \theta )}{1-\sin ^{2}\theta }\)
\(\frac{2\sin ^{2}\theta }{\cos ^{2}\theta }\)
2 tan2 θ
7 (b) \(2\tan ^{2}\theta =8\rightarrow \tan \theta =(\pm )2\)
( θ= ) =63.4°, 116.6°
Question
Solve the equation
\(\frac{tan \theta + 3 sin \theta + 2}{tan \theta – 3 sin \theta + 1\)
for \(0^o ≤ \theta ≤ 90^o\).
Answer/Explanation
Ans:
\(2tan \theta – 6 sin \theta + 2 = tan \theta + 3 sin \theta + 2 \rightarrow tan \theta – 9 sin \theta (=0)\)
\(sin \theta – 9 sin \theta cos \theta (=0)\)
\(sin \theta (1-9 cos \theta )(=0) \rightarrow sin \theta =0, cos \theta = \frac{1}{9}\)
\(\theta = 0\) or \(83.6^o\) (only answers in the given range)
Question.
(a). Prove the indentity \(\frac{1+sin\Theta }{cos\Theta }+\frac{cos\Theta }{1+sin\Theta}\equiv \frac{2}{cos\Theta }\)
(b) Hence solve the equation \(\frac{1+sin\Theta }{cos\Theta }+\frac{cos\Theta }{1+sin\Theta}\equiv \frac{2}{cos\Theta }\) for \(0 \leq \Theta \leq 2\Pi\).
Answer/Explanation
(a) \(\frac{(1+\sin \theta )^{2}+\cos ^{2}\theta }{\cos \theta (1+\sin \theta )}\)
Use of \(sin2 θ + cos2 θ = 1 \rightarrow \frac{2+2\sin \theta }{\cos \theta (1+\sin \theta )}\rightarrow \frac{2}{\cos \theta }\)
(b) \(\frac{2}{\cos \theta }=\frac{3}{\sin \theta }\rightarrow \tan \theta =1.5\)
θ = 0.983 or 4.12
(FT on second value for 1st value +π)