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

Guest Nov 13, 2014

#1**+5 **

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.

Melody
Nov 13, 2014

#2**+5 **

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)

CPhill
Nov 13, 2014

#3**0 **

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 :)

Guest Nov 13, 2014