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)\(\Theta \) is obtuse,\(\sin \Theta =k \)

\(\cos \Theta =-\sqrt{\left ( 1-k^{2} \right )}\)

(ii)\(\tan \Theta =\frac{\sin \Theta }{\cos \Theta }\)

\(\rightarrow \tan \Theta =-\frac{k}{\sqrt{\left ( 1-k^{2} \right )}}\)

(iii)\( \sin \left ( \Theta +\pi \right )=-k\)

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) 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 \(\frac{1-sin2x}{1+sin2x}=\frac{1}{3}\rightarrow sin2x=\frac{1}{2}
x=\frac{\Pi }{12}
x=\frac{5\Pi }{12}\)