What is $cos (arccos (\frac{3}{5}) - arcsin (\frac{4}{5}))$ $cos(arccos (3/5) - arcsin (4/5))$ ?

Question

8591 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-14T13:36:03

$1 .$ $1.$

Let, $arcsin (\frac{4}{5}) = α$ $arcsin(4/5)=alpha$ .

$:. sinalpha=4/5, where, alpha in [-pi/2,pi/2]=Q_1uuQ_4$ .

But, since $sinalpha=4/5 gt 0, alpha !in Q_4 :. alpha in Q_1$ .

Now, $cos^2alpha=1-sin^2alpha=1-(4/5)^2=9/25$ .

Knowing that, $alpha in Q_1, cosalpha=+sqrt(9/25)=+3/5$ .

Thus, $cosalpha=3/5, where, alpha in Q_1=[0,pi/2] sub [0,pi]$ .

$:." by definition, "arccos(3/5)=alpha=arcsin(4/5)$ .

$rArr (arccos(3/5)-arcsin(4/5))=0$ .

Consequently, $cos(arccos(3/5)-arcsin(4/5))=cos0=1$ .

$color(purple)("Enjoy Maths.!")$

Answer 2 · 2018-07-14T15:19:24

$1$

$cos((arccos)(3/5)-arc sine (4/5))$

$:.=cos($ (arccos(0.6)-arc sine (0.8))#

$:.=cos(53^@7'48''-53^@7'48'')$

$:.cos(0)=1$

What is cos(arccos (3/5) - arcsin (4/5))cos(arccos(35)−arcsin(45))cos(arccos (3/5) - arcsin (4/5))?