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Trigonometric Functions

Multiple Choice Questions

651.

$\cos (\frac{2 π}{7}) + \cos (\frac{4 π}{7}) + \cos (\frac{6 π}{7})$

is equal to zero
lies between 0 and 3
is a negative number
lies between 3 and 6

652.

The minimum value of $2^{\sin (x)} + 2^{\cos (x)}$ is

$2^{1 - 1 / \sqrt{2}}$
$2^{1 + 1 / \sqrt{2}}$
$2^{\sqrt{2}}$
2

653.
If $p = (\begin{matrix} \cos (\frac{π}{4}) & - \sin (\frac{π}{4}) \\ \sin (\frac{π}{4}) & \cos (\frac{π}{4}) \end{matrix}) and X = (\begin{matrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{matrix})$ . Then, p³X is equal to

$(\begin{matrix} 0 \\ 1 \end{matrix})$

$(\begin{matrix} - \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{matrix})$

$(\begin{matrix} - 1 \\ 0 \end{matrix})$

$(\begin{matrix} - \frac{1}{\sqrt{2}} \\ - \frac{1}{\sqrt{2}} \end{matrix})$

$(\begin{matrix} - 1 \\ 0 \end{matrix})$

$Given, P = (\begin{matrix} \cos (\frac{π}{4}) & - \sin (\frac{π}{4}) \\ \sin (\frac{π}{4}) & \cos (\frac{π}{4}) \end{matrix}) and X = (\begin{matrix} \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{matrix}) \Rightarrow P = \frac{1}{\sqrt{2}} (\begin{matrix} 1 & - 1 \\ 1 & 1 \end{matrix})$

$Now, P^{2} = P . P = \frac{1}{2} (\begin{matrix} 1 & - 1 \\ 1 & 1 \end{matrix}) . (\begin{matrix} 1 & - 1 \\ 1 & 1 \end{matrix}) = \frac{1}{2} (\begin{matrix} 1 - 1 & - 1 - 1 \\ 1 + 1 & - 1 + 1 \end{matrix}) = \frac{1}{2} (\begin{matrix} 0 & - 2 \\ 2 & 0 \end{matrix}) = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$

$P^{3} = P . P^{2} = \frac{1}{\sqrt{2}} (\begin{matrix} 1 & - 1 \\ 1 & 1 \end{matrix}) (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}) = \frac{1}{\sqrt{2}} (\begin{matrix} 0 - 1 & - 1 - 0 \\ 0 + 1 & - 1 + 0 \end{matrix}) = \frac{1}{\sqrt{2}} (\begin{matrix} - 1 & - 1 \\ 1 & - 1 \end{matrix})$

$Also given X = (\begin{matrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{matrix}) = \frac{1}{\sqrt{2}} (\begin{matrix} 1 \\ 1 \end{matrix}) ∴ P^{3} X = \frac{1}{\sqrt{2}} (\begin{matrix} - 1 & - 1 \\ 1 & - 1 \end{matrix}) . \frac{1}{\sqrt{2}} (\begin{matrix} 1 \\ 1 \end{matrix}) = \frac{1}{2} (\begin{matrix} - 1 & - 1 \\ 1 & - 1 \end{matrix}) = \frac{1}{2} (\begin{matrix} - 2 \\ 0 \end{matrix}) = (\begin{matrix} - 1 \\ 0 \end{matrix})$

654.

For $0 \leq P, Q \leq \frac{π}{2}$ , if $\sin (P) + \cos (Q) = 2$ , then the value of $\tan (\frac{P + Q}{2})$ is equal to

1
$\frac{1}{\sqrt{2}}$
$\frac{1}{2}$
$\frac{\sqrt{3}}{2}$

655.

The value of

$\cos^{2} (75 °) + \cos^{2} (45 °) + \cos^{2} (15 °) - \cos^{2} (30 °) - \cos^{2} (60 °)$ is

0
1
$\frac{1}{2}$
$\frac{1}{4}$

656.

The maximum and minimum values of $\cos^{6} (θ) + \sin^{6} (θ)$ are respectively

$1 and \frac{1}{4}$
1 and 0
2 and 0
1 and $\frac{1}{2}$

657.

Let $f (θ) = (1 + \sin^{2} (θ)) (2 - \sin^{2} (θ))$ . Then, for all values of $θ$

$f (θ) > \frac{9}{4}$
$f (θ) < 2$
$f (θ) > \frac{11}{4}$
$2 \leq f (θ) \leq \frac{9}{4}$

658.

If P, Q and R are angles of an isosceles triangle and $∠ P = \frac{π}{2}$ , then the value of

${(\cos (\frac{P}{3}) - isin (\frac{P}{3}))}^{3} + (\cos (Q) + isin (Q)) (\cos (R) - isin (R)) + (\cos (P) - isin (P)) (\cos (Q) - isin (Q)) (\cos (R) - isin (R))$