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I thought using sin(A+B) and Cos(A+B) would work but I got no where.

Solve these equations for 0 < θ < 360o, giving θ to 1 decimal place where appropriate:

(a) sin(p+15degrees)= 3cos(p+15degrees)

(b) sinpcosp= 1/2

Guest Nov 15, 2014

#1**+13 **

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

CPhill Nov 15, 2014

#1**+13 **

Best Answer

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

CPhill Nov 15, 2014

#2**+10 **

Slightly simpler way of doing (a) is to divide both sides by cos(p+15°) to get tan(p+15°) = 3

p+15° = tan^{-1}(3)

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\mathtt{3}}\right)} = {\mathtt{71.565\: \!051\: \!177\: \!078^{\circ}}}$$

so p = 71.565° - 15°

or p = 56.6° (to one decimal place)

That's the first quadrant solution. Add 180° to get the third quadrant solution.

.

Alan Nov 16, 2014