CPhill

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 #1
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The second isn't too bad..we have

 (b)  sinpcosp= 1/2

Note that sinpcosp is really just (1/2)(2)sinpcosp   = (1/2)sin(2p)

So we have

(1/2)sin(2p) = 1/2        multiply through by 2

sin(2p) = 1

Since sinp = 1 at  90 and 450, then sin(2p) = 1 at 45 and 225

Here's the graph of this one......https://www.desmos.com/calculator/f1wb8idmiy

 

For a, we have

sin(p+15degrees)= 3cos(p+15degrees)   ...note by an identity sinA = cos(90-A)

So......let A = p + 15   and we have

sin(p+ 15) = cos(90- (p+ 15)) = cos (75 - p)....so

cos(75 - p)  = 3cos(15 + p)...and we have

cos75cosp + sin75sinp = 3[cos15cosp - sin15 sinp]  rearranging, we have

3cos15cosp - cos75cosp =  sin75sinp + 3sin15sinp

cosp(3cos15-cos75) = sinp(sin75 + 3sin15)    rearrange again

sinp / cos p = (3cos15 - cos75)/(sin75 + 3 sin15)

tanp = (3cos15 - cos75)/(sin75 + 3 sin15)

tan-1 (3cos15 - cos75)/(sin75 + 3 sin15) = p = 56.56505117705 degrees...this could also be a 3rd quad angle = (180 + 56.56505117705) degrees = 236.56505117705 degrees

This one was definitely tougher !!

Here's the graph .....  https://www.desmos.com/calculator/8qdesoejgk

 

Nov 15, 2014