cos (sin^-1 1/4 - tan^-1 (-3) ) Note: I posted a picture below with my work and i wanted to see if i'm correct, if not, please help me out, thanks!!! <3
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cosx =1/sqrt15 cosy=1/sqrt 10 sinx=1/4 siny=-3/sqrt10 nw apply cos(x-y) ,u get 1/sqrt150 - 3/4sqrt10
Thanks, can you help me out with my other post? =P
which one will try
Solve the following equation for solutions over the interval [0 degrees, 360 degrees)
Hello, boltage619! Factor: . We have two equations to solve . . . . .
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