First slide

Multiple and sub- multiple Angles

Question

A line OA of length r starts from its initial position OX and traces an angle $AOB = α$ in the anticlockwise direction. It then traces back in the clockwise direction at an angle BOC - 30 (where $α > 3 θ)$ . Z is the foot of the perpendicular form C on OA. Also, $\frac{\sin^{3} θ}{CL} = \frac{\cos^{3} θ}{OL} = 1$ .

NA

Question

$\frac{1 - rcos α}{rsin α}$ is equal to

A

$\tan 2 θ$
B

$\cot 2 θ$
C

$\sin 2 θ$
D

$\cos 2 θ$

Solution

$\begin{array}{l} OL = rcos (α - 3 θ) \\ CL = rsin (α - 3 θ) \end{array}$
So, $\frac{\sin^{3} θ}{rsin (α - 3 θ)} = \frac{\cos^{3} θ}{rcos (α - 3 θ)} = 1$
$\Rightarrow \frac{\sin^{3} θ}{\sin (α - 3 θ)} = \frac{\cos^{3} θ}{\cos (α - 3 θ)} = r$
Now $\cos^{4} θ - \sin^{4} θ$
$\begin{array}{l} = \cos θ (rcos (α - 3 θ)) - \sin θ (rsin (α - 3 θ)) \\ \Rightarrow \cos 2 θ = rcos (α - 2 θ) \\ \Rightarrow \cos 2 θ = r (\cos αcos 2 θ + \sin αsin 2 θ) \\ \Rightarrow (1 - rcos α) \cos 2 θ = rsin αsin 2 θ \\ \Rightarrow \frac{1 - rcos α}{rsin α} = \tan 2 θ . . . . . (i) \\ \cos^{3} θsin θ + \sin^{3} θcos θ \\ \Rightarrow \sin θcos θ = rsin (α - 2 θ) \\ \Rightarrow \sin 2 θ = 2 r [\sin αcos 2 θ - \cos αsin 2 θ] \\ \Rightarrow (1 + 2 rcos α) \sin 2 θ = 2 rsin αcos 2 θ \\ \Rightarrow \frac{1 + 2 rcos α}{2 rsin α} = \cot 2 θ . . . . . (ii) \end{array}$
From (1) and (2),we get
$\begin{array}{l} \frac{1 - rcos α}{rsin α} = \frac{2 rsin α}{1 + 2 rcos α} \\ \Rightarrow 1 - rcos α + 2 rcos α - 2 r^{2} \cos^{2} α = 2 r^{2} \sin^{2} α \\ \Rightarrow \frac{2 r^{2} - 1}{r} = \cos α \end{array}$

Question

$\frac{2 rsin α}{1 + 2 rcos α}$ is equal to

A

$\tan^{2} θ$
B

$\cot^{2} θ$
C

$\cot 2 θ$
D

$\tan 2 θ$

Solution

$\begin{array}{l} \cos^{4} θ - \sin^{4} θ = \cos θ (rcos (α - 3 θ)) - \sin θ (rsin (α - 3 θ)) \\ \Rightarrow \cos 2 θ = rcos (α - 2 θ) \\ \Rightarrow \cos 2 θ = r (\cos αcos 2 θ + \sin αsin 2 θ) \\ \Rightarrow (1 - rcos α) \cos 2 θ = rsin αsin 2 θ \\ \Rightarrow \frac{1 - rcos α}{rsin α} = \tan 2 θ . . . . . (i) \\ \cos^{3} θsin θ + \sin^{3} θcos θ \\ \Rightarrow \sin θcos θ = rsin (α - 2 θ) \\ \Rightarrow \sin 2 θ = 2 r [\sin αcos 2 θ - \cos αsin 2 θ] \\ \Rightarrow (1 + 2 rcos α) \sin 2 θ = 2 rsin αcos 2 θ \\ \Rightarrow \frac{1 + 2 rcos α}{2 rsin α} = \cot 2 θ . . . . . (ii) \end{array}$

Question

$\frac{2 r^{2} - 1}{r}$ is equal to

A

$\sin α$
B

$\cos α$
C

$\sin θ$
D

$\cos θ$

Solution

$\begin{array}{l} \cos^{4} θ - \sin^{4} θ = \cos θ (rcos (α - 3 θ)) - \sin θ (rsin (α - 3 θ)) \\ \Rightarrow \cos 2 θ = rcos (α - 2 θ) \\ \Rightarrow \cos 2 θ = r (\cos αcos 2 θ + \sin αsin 2 θ) \\ \Rightarrow (1 - rcos α) \cos 2 θ = rsin αsin 2 θ \\ \Rightarrow \frac{1 - rcos α}{rsin α} = \tan 2 θ . . . . . (i) \\ \cos^{3} θsin θ + \sin^{3} θcos θ \\ \Rightarrow \sin θcos θ = rsin (α - 2 θ) \\ \Rightarrow \sin 2 θ = 2 r [\sin αcos 2 θ - \cos αsin 2 θ] \\ \Rightarrow (1 + 2 rcos α) \sin 2 θ = 2 rsin αcos 2 θ \\ \Rightarrow \frac{1 + 2 rcos α}{2 rsin α} = \cot 2 θ . . . . . (ii) \end{array}$
from (i) and (ii)
$\begin{array}{l} \frac{1 - rcos α}{rsin α} = \frac{2 rsin α}{1 + 2 rcos α} \\ \Rightarrow 1 - rcos α + 2 rcos α - 2 r^{2} \cos^{2} α = 2 r^{2} \sin^{2} α \\ \Rightarrow \frac{2 r^{2} - 1}{r} = \cos α \end{array}$

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