Evalulate Basic Problems

Question

My professor said for this problem to use inverse trig functions, half angle, and double angle identities.

 

Evaluate:

 

tan(2sin^-1(4/5))

 

Please list the answer step by step and next to each step please include the identity used! :)

Answer 1

Hi Kailes,

 

One thing to realize is that if an angle has a sine equal to 4/5, then it has a tangent of 4/3. This is because we are dealing with a 3,4,5 right triangle. You will see why this is mentioned in the next paragraph.

 

The trig identity for tangent of twice an angle (double angle formula for tangent):

 

tan 2u = 2tanu/(1 - tan²u)

 

So in effect we are looking for 2tan(of the angle that has a sine of 4/5)/(1 - tan²(of that same angle)).

 

That is: tan(2arcsin(4/5)) = 2(4/3)/(1 - (4/3)²) = 8/3 / (1 - 16/9) = 24/-7 = -3.428571...

 

Actually, there are infinitely many angles that have a sine = 4/5. One other is in the 2nd quadrant where the tangent of that angle would be -4/3. 

Answer 2

tan(2sin-1(4/5))=tan(2sin-(2*2)/2(2)+1)=tan(2sin-sin 30(2*2)/3)=tan(2sin-(2*2)/sin30(2+1))=tan(2sin-1(2*2)/sin 30*(3*3)(sq.root)=tan(2sin-1(2*2)sin30/cos 30 cos 30)=tan(2sin-1*sin30/1/2*1/2cos 30 cos 30)=tan(2sin-1sin30/sin30*1/2cos 30 cos 30)=tan(2sin-1/2cos 30 cos 30)=tan(2sin-1/2cos60 cos 60)=tan(2sin-sec60/sec60 cos 60)=tan(2sin-sec60)=tan(2sin-2)=tan(2sin(1)-sin(inverse)1)tan(2*pie/2-pie/2)=tan(pie/2)=infinite