Question
Given that \(\Theta \) is an obtuse angle measured in radians and that \(\sin \Theta =k\),find ,in terms of k ,an expression for
(i)\(\cos \Theta \)
(ii)\(\tan \Theta \)
(iii)\(\sin \left ( \Theta +\pi \right )\)
▶️Answer/Explanation
(i)\(\sin^2 \Theta + \cos^2 \Theta = 1\)
\( k^2 + \cos^2 \Theta = 1 \)
\( \cos^2 \Theta = 1 – k^2 \)
Since \(\Theta\) is in the second quadrant, \(\cos \Theta\) is negative. Therefore:
\( \cos \Theta = -\sqrt{1 – k^2} \)
(ii) \( \tan \Theta = \frac{\sin \Theta}{\cos \Theta} = \frac{k}{-\sqrt{1 – k^2}} \)
(iii) \( \sin (\Theta + \pi) = \sin \Theta \cos \pi + \cos \Theta \sin \pi \) (\(\sin (A + B) = \sin A \cos B + \cos A \sin B\))
\( \sin (\Theta + \pi) = k(-1) + 0 \)
\( \sin (\Theta + \pi) = -k \)
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)Solving LHS
\(\left(\frac{1}{\cos x} – \tan x\right)^2\)
\(= \left(\frac{1}{\cos x} – \frac{\sin x}{\cos x}\right)^2\) (since \(\tan x = \frac{\sin x}{\cos x}\))
\(= \left(\frac{1 – \sin x}{\cos x}\right)^2\)
\(= \frac{(1 – \sin x)^2}{\cos^2 x}\)
\(= \frac{1 – 2\sin x + \sin^2 x}{\cos^2 x}\)
\(= \frac{1 – 2\sin x + \sin^2 x}{1 – \sin^2 x}\)
\(= \frac{(1 – \sin x)^2}{(1 + \sin x)(1 – \sin x)}\) (factorizing the denominator)
\(= \frac{1 – \sin x}{1 + \sin x}\)
(ii)Substituting \(\tan x\) with \(\tan 2x\) in the identity:
\(\left(\frac{1}{\cos x} – \tan 2x\right)^2 = \frac{1 – \sin x}{1 + \sin x}\)
\(\frac{1 – \sin x}{1 + \sin x} = \frac{1}{3}\)
\(3(1 – \sin x) = 1 + \sin x\)
\(3 – 3\sin x = 1 + \sin x\)
\(4\sin x = 2\)
\(\sin x = \frac{1}{2}\)
In the given range \(0 \leq x \leq 2\pi\) where \(\sin x = \frac{1}{2}\).
The solutions are
\(x = \frac{\pi}{6}, \frac{5\pi}{6}\)