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)

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