# How to simplify this Cosecant identities?

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, 8 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

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 $$\cos (2A) = 2 \cos^2 (A) - 1$$

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

$$\large = 3 \times 2 = 6$$

- 4 years, 8 months ago