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 a

You are amazing!! Thank you so much! :))) I hope you have a wonderful evening! 

Let's see

both those things in the brackets are angles.

so the question becomes sin(A+B)=sinACosB+cosAsinB

Now 

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.

$$\\sin(A+B)\\\\ =sinACosB+cosAsinB\\\\ =\frac{1}{4}\times\frac{1}{\sqrt{26}}+\frac{\sqrt{15}}{4}\times\frac{-5}{\sqrt26}\\\\ =\frac{1}{4\sqrt{26}}+\frac{-5\sqrt{15}}{4\sqrt{26}}\\\\ =\frac{1-5\sqrt{15}}{4\sqrt{26}}\\\\ =\frac{\sqrt{26}\left(1-5\sqrt{15}\right)}{104}\\\\$$

(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

V=cos2(pi)t

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

.4=cos2(pi)t  

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