+0

# solving these mindboggling trig stuff

0
316
4

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
+80865
+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
Sort:

#1
+80865
+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
#2
+26397
+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)

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

so p = 71.565° - 15°

or p = 56.6° (to one decimal place)

.

Alan  Nov 16, 2014
#3
+80865
+5

That ain't "slightly simpler".....that's WAY more simple.....thanks, Alan!!!.....(Melody's "Method of Over-Complication" is rubbing off on me....!!!!)

CPhill  Nov 16, 2014
#4
+91432
0

Yes that proves it - I'm a great teacher!!     LOL

Melody  Nov 17, 2014

### 15 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details