can you explain how to solve sin(sin^-1(1/4)+tan^-1(-5)) i have the answer but i dont know how to get to it. I need it in radical form not as a decimal
both those things in the brackets are angles.
so the question becomes sin(A+B)=sinACosB+cosAsinB
It needs to be remembered that
the range for both inverse sine and inverse tan is from -pi/2 to pi/2 inclusive
A=sin^-1(1/4) so A is in the 1st quad sinA=1/4 and cosA=sqrt(15 )/4
B=tan^-1(-5) so B is in the 4th quad tanB=-5 and sinB=-5/sqrt(26) and cosB=+1/sqrt(26)
I got these ratios by drawing the right angled triangles for A and B and finig the third side use Pythagoras' theorm.
(I have edited this a little as I made a mistake)
If you want me to explain anything just ask but think about it a bit first :)
Here are the triangles that I used.
sin(sin^-1(1/4)+tan^-1(-5)) Let sin-1(1/4) = A and let tan-1(-5) = B
So we have
sin(A + B) = sinAcosB + sinBcosA
Note that, since the tan-1 is negative, B must lie in the 4th quadrant. So the cosB = 1/√26 and sinB = -5/√26. And cosA = √15/4......so we have
sin(A + B) = (1/4)(1/√26) + ( -5/√26) (√15/4.) = (1 - 5√15)/ (4√26)
Thank you guys! I was able to get the right answer with your help! :)
I have one other question for you!
the question is: Let the equation below model the electromotive fore in volts V at t seconds
find the least value of t where t is less than or equal to 1/2 and greater than or equal to 0 for V=0.4
I know the answer to this is t=0.18 however i dont know how they got that answer!
Thanks again for your help :)
So we have
Using the cosine inverse, we have
cos-1(2pi(t)) = .4 let (2pi)t = x
cos-1(x) = .4
x = 1.159
So x = (2pi)t = 1.159 divide both sides by 2pi
t = 1.159/ (2pi) = about 0.184