find exact value when defined: cos[arctan(-3/4)-arcsin(4/5)]?

Question

value is 0. Please explain.

its cos[arctan(-3/4)-arcsin(4/5)]

thanks anak i just got the sign of sin ^^ mixed up.

Answer 1

Ok, let's go on :

cos[arctan(-3/4) - arcsin(4/5)]

Let's set : alpha = arctan(-3/4)

beta = arcsin(4/5)

cos[alpha - beta] :

cos(alpha)*cos(beta) + sin(alpha)*sin(beta)

But : tan(alpha) = -3 / 4 >>>> cos(alpha) = 4 / 5

sin(alpha) = -3/5

sin(beta) = 4/5 >>>>> cos(beta) = 3/5

replacing on :

cos(alpha)*cos(beta) + sin(alpha)*sin(beta)

4/5*3/5 - 3/5*4/5 = 0

Hope that helps

Answer 2

You have a 3,4,5 triangle situation, but in different quadrants. arctan(-3/4) =180-36= 144 deg. The arcsin(4/5)= 54 deg. The difference is 90 deg. The Cos 90=0, q.e.d.

Answer 3

have to use cos^2+sin^2 =1 cos^2 (2x) = 1-sin^2 (2x) = 1- (-3/4)^2 = 1-9/16 = 7/16 taking sq rt cos (2x) = sq rt(7)/4