Trigonometric Ratio and Identities MCQ Questions & Answers in Trigonometry | Maths

Use $$\sin 2\alpha = \sin \left\{ {\left( {\alpha + \beta } \right) + \left( {\alpha - \beta } \right)} \right\}$$
$$\sin 2\alpha = \sin \left( {\alpha + \beta } \right) \cdot \cos \left( {\alpha - \beta } \right) + \cos \left( {\alpha + \beta } \right) \cdot \sin \left( {\alpha - \beta } \right).$$

We know that $$\tan \theta = \cot \theta - 2\cot 2\theta .$$
∴ the expression $$ = \left( {\cot \frac{\pi }{{16}} - 2\cot \frac{\pi }{8}} \right) + 2\left( {\cot \frac{\pi }{8} - 2\cot \frac{\pi }{4}} \right) + 4 = \cot \frac{\pi }{{16}}.$$

$$\eqalign{ & a = d\cos \alpha + c\sin \alpha ,b = c\cos \alpha - d\sin \alpha \cr & \Rightarrow \,\,{a^2} + {b^2} = {c^2} + {d^2}. \cr} $$

Let the angles are $$\alpha $$ and $$\beta ,$$ then $$\alpha - \beta = {1^ \circ }$$
$$ \Rightarrow \alpha - \beta = \frac{\pi }{{{{180}^ \circ }}}$$    is circular measure $$\,\,\,\,.....\left( {\text{i}} \right)$$
As given, $$\alpha + \beta = 1\,\,\,\,\,.....\left( {{\text{ii}}} \right)$$
On solving Eqs. (i) and (ii), we get,
$$\alpha = \frac{1}{2}\left[ {1 + \frac{\pi }{{{{180}^ \circ }}}} \right]{\text{ and }}\beta = \frac{1}{2}\left[ {1 - \frac{\pi }{{{{180}^ \circ }}}} \right]$$
$$\beta $$ is the smaller angle.
Hence, smaller angle $$ = \frac{1}{2}\left[ {1 - \frac{\pi }{{{{180}^ \circ }}}} \right]$$

The given expression
$$\eqalign{ & = {\left( {\frac{{2\cos \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{A - B}}{2}} \right)}}{{2\cos \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{A - B}}{2}} \right)}}} \right)^n} + {\left( {\frac{{2\sin \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{A - B}}{2}} \right)}}{{2\sin \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{B - A}}{2}} \right)}}} \right)^n} \cr & = {\cot ^n}\left( {\frac{{A - B}}{2}} \right) + {\left( { - 1} \right)^n}{\cot ^n}\left( {\frac{{A - B}}{2}} \right) \cr} $$
\[ = \left\{ \begin{array}{l} 2\,{\cot ^n}\left( {\frac{{A - B}}{2}} \right),{\rm{ if }}\,n\,{\rm{ is\, even}}\\ 0,{\rm{ if }}\,n\,{\rm{ is\, odd}} \end{array} \right.\]

The expression $$ = \tan \theta \cdot \frac{{\sqrt 3 + \tan \theta }}{{1 - \sqrt 3 \tan \theta }} \cdot \frac{{\sqrt 3 - \tan \theta }}{{1 + \sqrt 3 \tan \theta }}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }} = \tan 3\theta .$$

$$\eqalign{ & 3{\left( {\sin x - \cos x} \right)^4} + 6{\left( {\sin x + \cos x} \right)^2} + 4\left( {{{\sin }^6}x + {{\cos }^6}x} \right) \cr & = 3{\left( {1 - \sin 2x} \right)^2} + 6\left( {1 + \sin 2x} \right) + 4\left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^3} - 3{{\sin }^2}x{{\cos }^2}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)} \right] \cr & = 3 - 6\sin 2x + 3{\sin ^2}2x + 6 + 6\sin 2x + 4\left[ {1 - \frac{3}{4}{{\sin }^2}2x} \right] \cr & = 13 + 3{\sin ^2}2x - 3{\sin ^2}2x \cr & = 13 \cr} $$

$$\eqalign{ & f\left( x \right) = \left( {\sin {x^7}} \right) \cdot {e^{{x^5}\operatorname{sgn} {x^9}}} \cr & \sin {x^7} \to {\text{odd,}}\operatorname{sgn} {x^9} \to {\text{odd, }}{x^5} \to {\text{odd}} \cr & \therefore {e^{{x^5}\operatorname{sgn} {x^9}}} \to {\text{even}} \cr & \therefore f\left( x \right) = {\text{odd}} \times {\text{even}} = {\text{odd}} \cr} $$

Use the fact that opposite angles are supplementary.

Here, $$2\sin \frac{{\alpha + \beta }}{2} \cdot \cos \frac{{\alpha - \beta }}{2} = a,2\sin \frac{{\alpha + \beta }}{2} \cdot \sin \frac{{\beta - \alpha }}{2} = b.$$
Now, divide and get the value.