# What is the value of cos [arc cos (3/5) + arc sin (4/5)]?

*print*Print*list*Cite

### 2 Answers

We have to find the value of cos [arc cos (3/5) + arc sin (4/5)].

Now arc cos (3/5) and arc sin (4/5) are two angles. Therefore each of them has a value for the sine and cosine function.

We know that cos (x + y) = cos x * cos y – sin x * sin y

Also, arc cos (3/5) can be written as arc sin [sqrt (1 – (3/5) ^2)]

And arc sin (4 /5) can be written as arc cos [sqrt (1 – (4/5) ^2)]

So cos [arc cos (3/5) + arc sin (4/5)]

=> cos [arc cos (3/5)] * cos [arc sin (4/5)] - sin [arc cos (3/5)] * sin [arc sin (4/5)]

=> 3/5 * cos [arc cos (sqrt (1 – (4/5) ^2))] – sin [arc sin (sqrt (1 – (3/5) ^2))]* (4/5)

=> (3/5)* sqrt (1 – (4/5) ^2) – (sqrt 1- (3/5) ^2) * (4/5)

=> (3/5)* sqrt (9 / 25) – sqrt (16/25)* (4/5)

=> (3/5)*(3/5) – (4/5)*(4/5)

=> 9/25 – 16/25

=> -7/25

**Therefore cos [arc cos (3/5) + arc sin (4/5)] = -7/25**

What is the value of cos [arc cos (3/5) + arc sin (4/5)].

Cos(A+B) = cosAcosB-sinAsinB

cos(arccosa) = a.

sin(arcsina) = a

cos(arcsinx) = sqrt(1-x^2).

sin(arccosy) = sqrt(1-x^2).

Therefore,

cos [arc cos (3/5) + arc sin (4/5)] = (3/5)*sqrt{1-(4/5)^2}- (4/5)*sqrt{1-(3/5)^2}.

cos [arc cos (3/5) + arc sin (4/5)] = 3^/25-4^2/25.

cos [arc cos (3/5) + arc sin (4/5)] = (9-25)/25.

cos [arc cos (3/5) + arc sin (4/5)] = -7/25.

cos [arc cos (3/5) + arc sin (4/5)] = -0.28.