Question #01a2a

Question

Question #01a2a

1 Answer

Impact of this question

1014 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-30T07:58:54

$\frac{1}{2} (cos (5 θ) - cos (9 θ))$ $1/2(cos(5theta)-cos(9theta))$

Explanation:

Using the sum and difference formulas for cosine we have:

$cos (x - y) = cos (x) cos (y) + sin (x) sin (y)$ $cos(x-y)=cos(x)cos(y)+sin(x)sin(y)$
$cos (x + y) = cos (x) cos (y) - sin (x) sin (y)$ $cos(x+y)=cos(x)cos(y)-sin(x)sin(y)$

If we subtract those formulas:

$cos (x - y) - cos (x + y) = 2 sin (x) sin (y)$ $cos(x-y)-cos(x+y) =2sin(x)sin(y)$

so

$sin (x) sin (y) = \frac{1}{2} (cos (x - y) - cos (x + y))$ $sin(x)sin(y) = 1/2(cos(x-y)-cos(x+y))$

so

$sin (2 θ) sin (7 θ) = \frac{1}{2} (cos (2 θ - 7 θ) - cos (2 θ + 7 θ))$ $sin(2theta)sin(7theta) = 1/2(cos(2theta-7theta)-cos(2theta+7theta))$

$= \frac{1}{2} (cos (- 5 θ) - cos (9 θ))$ $=1/2(cos(-5theta)-cos(9theta))$

since cosine is even we can rewrite this as:

$= \frac{1}{2} (cos (5 θ) - cos (9 θ))$ $=1/2(cos(5theta)-cos(9theta))$

Science

Math

Humanities

... and beyond

Question #01a2a

1 Answer

Explanation:

Related questions

Impact of this question