How do you evaluate $sin [arcsin (\frac{3}{5}) - arccos (\frac{3}{5})]$ $sin[arcsin(3/5)-arccos(3/5) ]$ ?

Question

How do you evaluate $sin [arcsin (\frac{3}{5}) - arccos (\frac{3}{5})]$ $sin[arcsin(3/5)-arccos(3/5) ]$ ?

1 Answer

Impact of this question

10528 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-15T18:15:09

$- \frac{7}{25}$ $-7/25$

Explanation:

Let $θ = a r c sin (\frac{3}{5}) . sin θ = \frac{3}{5}$ $theta = arc sin(3/5). sin theta = 3/5$ .
Restricting to the principal value,
$a r c cos (\frac{3}{5}) = \frac{π}{2} - a r c sin (\frac{3}{5}) = \frac{π}{2} - θ$ $arc cos(3/5)=pi/2-arc sin(3/5)= pi/2-theta$ .

$sin (arcsin (\frac{3}{5}) - a r c cos (\frac{3}{5})) = sin (θ - (\frac{π}{2} - θ)) = - sin (\frac{π}{2} - 2 θ) = - cos 2 θ = - (1 - 2 {sin}^{2} θ) = - (1 - 2 {(\frac{3}{5})}^{2}) = - \frac{7}{25}$ $sin(arcsin (3/5)-arc cos (3/5)) =sin(theta-(pi/2-theta))=-sin(pi/2-2theta)=-cos 2theta=-(1-2 sin^2theta)=-(1-2(3/5)^2)=-7/25$

Science

Math

Humanities

... and beyond

How do you evaluate $sin [arcsin (\frac{3}{5}) - arccos (\frac{3}{5})]$ $sin[arcsin(3/5)-arccos(3/5) ]$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you evaluate sin[arcsin(35)−arccos(35)]sin[arcsin(3/5)-arccos(3/5) ]?

1 Answer

Explanation:

Related questions

Impact of this question

How do you evaluate $sin [arcsin (\frac{3}{5}) - arccos (\frac{3}{5})]$ $sin[arcsin(3/5)-arccos(3/5) ]$ ?