Question

What is $${\sin ^2}66\frac{{{1^ \circ }}}{2} - {\sin ^2}23\frac{{{1^ \circ }}}{2}$$     equal to ?

A. $$\sin {47^ \circ }$$
B. $$\cos {47^ \circ }$$  
C. $$2\sin {47^ \circ }$$
D. $$2\cos {47^ \circ }$$
Answer :   $$\cos {47^ \circ }$$
Solution :
$$\eqalign{ & {\sin ^2}66\frac{{{1^ \circ }}}{2} - {\sin ^2}23\frac{{{1^ \circ }}}{2} \cr & = {\left[ {\sin \left( {{{90}^ \circ } - 23\frac{{{1^ \circ }}}{2}} \right)} \right]^2} - {\sin ^2}23\frac{{{1^ \circ }}}{2} \cr & = {\cos ^2}23\frac{{{1^ \circ }}}{2} - {\sin ^2}23\frac{{{1^ \circ }}}{2} \cr & = \cos 2\left( {23\frac{{{1^ \circ }}}{2}} \right) = \cos {47^ \circ }\,\,\left( {\because \cos 2A = {{\cos }^2}A - {{\sin }^2}A} \right) \cr & = \cos \left[ {2 \times \left( {\frac{{47}}{2}} \right)} \right] = \cos {47^ \circ } \cr} $$

Releted MCQ Question on
Trigonometry >> Trigonometric Ratio and Identities

Releted Question 1

If $$\tan \theta = - \frac{4}{3},$$   then $$\sin \theta $$  is

A. $$ - \frac{4}{5}{\text{ but not }}\frac{4}{5}$$
B. $$ - \frac{4}{5}{\text{ or }}\frac{4}{5}$$
C. $$ \frac{4}{5}{\text{ but not }} - \frac{4}{5}$$
D. None of these
Releted Question 2

If $$\alpha + \beta + \gamma = 2\pi ,$$    then

A. $$\tan \frac{\alpha }{2} + \tan \frac{ \beta }{2} + \tan \frac{\gamma }{2} = \tan \frac{\alpha }{2}\tan \frac{\beta }{2}\tan \frac{\gamma }{2}$$
B. $$\tan \frac{\alpha }{2}\tan \frac{\beta }{2} + \tan \frac{\beta }{2}\tan \frac{\gamma }{2} + \tan \frac{\gamma }{2}\tan \frac{\alpha }{2} = 1$$
C. $$\tan \frac{\alpha }{2} + \tan \frac{ \beta }{2} + \tan \frac{\gamma }{2} = - \tan \frac{\alpha }{2}\tan \frac{\beta }{2}\tan \frac{\gamma }{2}$$
D. None of these
Releted Question 3

Given $$A = {\sin ^2}\theta + {\cos ^4}\theta $$    then for all real values of $$\theta $$

A. $$1 \leqslant A \leqslant 2$$
B. $$\frac{3}{4} \leqslant A \leqslant 1$$
C. $$\frac{13}{16} \leqslant A \leqslant 1$$
D. $$\frac{3}{4} \leqslant A \leqslant \frac{{13}}{{16}}$$
Releted Question 4

The value of the expression $$\sqrt 3 \,{\text{cosec}}\,{\text{2}}{{\text{0}}^ \circ } - \sec {20^ \circ }$$     is equal to

A. 2
B. $$\frac{{2\sin {{20}^ \circ }}}{{\sin {{40}^ \circ }}}$$
C. 4
D. $$\frac{{4\sin {{20}^ \circ }}}{{\sin {{40}^ \circ }}}$$

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