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How to find the value of this equation? I've tried about the multiplication and addition of sine and cosine, but it still useless, and I always ended up with the sin 10 at the end.

Note by Leonardo Chandra 4 years ago

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2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

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Recall the identities

\( \sin (A) + \sin (B) = 2 \sin ( \frac {A+B}{2} ) \space \cos ( \frac {A-B}{2} ) \)

\( - \cos (A) + \cos (B) = 2 \sin ( \frac {A+B}{2} ) \space \sin ( \frac {A-B}{2} ) \)

\( \cos (A) + \cos (B) = 2 \cos ( \frac {A+B}{2} ) \space \cos ( \frac {A-B}{2} ) \)

We have,

\( \large \frac {1}{\sin (10^\circ )} + \frac {1}{\sin (50^\circ )} - \frac {1}{\sin (70^\circ )}\)

\( \large = \frac {1}{\sin (10^\circ )} + \frac {1}{\sin (50^\circ )} - \frac {1}{\cos (20^\circ )}\), because \( \sin (A) = \cos (90^\circ - A) \)

\( \large = \frac {2 ( \space \sin (50^\circ ) + \sin (10^\circ ) \space ) } { 2 \sin (50^\circ ) \space \sin (10^\circ ) } - \frac {1}{\cos (20^\circ )} \)

\( \large = \frac {2 (2 \sin (30^\circ) \cos (20^\circ) ) } { - \cos (60^\circ) + \cos (40^\circ) } - \frac {1}{\cos (20^\circ )} \)

\( \large = \frac {4 \cos (20^\circ) } { - 1 + 2\cos (40^\circ) } - \frac {1}{\cos (20^\circ )} \)

\( \large = \frac {4 \cos^2 (20^\circ) -2 \cos (40^\circ) + 1} { 2 \cos (40^\circ) \space \cos (20^\circ) - \cos (20^\circ) } \)

\( \large = \frac {4 \cos^2 (20^\circ) - 2 -2 \cos (40^\circ) + 3} { ( \space \cos (60^\circ) + \cos (20^\circ) ) - \cos (20^\circ) } \)

\( \large = \frac { 2( 2 \cos^2 (20^\circ) - 1 ) -2 \cos (40^\circ) + 3} { \cos (60^\circ) } \), because we want to apply \( \cos (2A) = 2 \cos^2 (A) - 1 \)

\( \large = \frac { 2 \cos (40^\circ) -2 \cos (40^\circ) + 3} { \frac {1}{2} } \)

\( \large = 3 \times 2 = 6 \)

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`*italics*`

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italics`**bold**`

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boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

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Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

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`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestRecall the identities

\( \sin (A) + \sin (B) = 2 \sin ( \frac {A+B}{2} ) \space \cos ( \frac {A-B}{2} ) \)

\( - \cos (A) + \cos (B) = 2 \sin ( \frac {A+B}{2} ) \space \sin ( \frac {A-B}{2} ) \)

\( \cos (A) + \cos (B) = 2 \cos ( \frac {A+B}{2} ) \space \cos ( \frac {A-B}{2} ) \)

We have,

\( \large \frac {1}{\sin (10^\circ )} + \frac {1}{\sin (50^\circ )} - \frac {1}{\sin (70^\circ )}\)

\( \large = \frac {1}{\sin (10^\circ )} + \frac {1}{\sin (50^\circ )} - \frac {1}{\cos (20^\circ )}\), because \( \sin (A) = \cos (90^\circ - A) \)

\( \large = \frac {2 ( \space \sin (50^\circ ) + \sin (10^\circ ) \space ) } { 2 \sin (50^\circ ) \space \sin (10^\circ ) } - \frac {1}{\cos (20^\circ )} \)

\( \large = \frac {2 (2 \sin (30^\circ) \cos (20^\circ) ) } { - \cos (60^\circ) + \cos (40^\circ) } - \frac {1}{\cos (20^\circ )} \)

\( \large = \frac {4 \cos (20^\circ) } { - 1 + 2\cos (40^\circ) } - \frac {1}{\cos (20^\circ )} \)

\( \large = \frac {4 \cos^2 (20^\circ) -2 \cos (40^\circ) + 1} { 2 \cos (40^\circ) \space \cos (20^\circ) - \cos (20^\circ) } \)

\( \large = \frac {4 \cos^2 (20^\circ) - 2 -2 \cos (40^\circ) + 3} { ( \space \cos (60^\circ) + \cos (20^\circ) ) - \cos (20^\circ) } \)

\( \large = \frac { 2( 2 \cos^2 (20^\circ) - 1 ) -2 \cos (40^\circ) + 3} { \cos (60^\circ) } \), because we want to apply \( \cos (2A) = 2 \cos^2 (A) - 1 \)

\( \large = \frac { 2 \cos (40^\circ) -2 \cos (40^\circ) + 3} { \frac {1}{2} } \)

\( \large = 3 \times 2 = 6 \)

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