# How do i solve this trig question? identities etc?

1. Mar 20, 2009

### Dell

How do i solve this trig question? identities etc?
4*[cos(B) + 3*sin(B)]=1+ 2*[3*sin(39) - cos(39)]

i can get it to
cosB + 3sinB=0.08054076 using basic algebra, but how do i find a value for B

2. Mar 20, 2009

### danago

Can you think of a way to perhaps combine the sine and cosine terms of the left into a single trig function?

3. Mar 20, 2009

### Dell

no, thats what i need to somehow do

4. Mar 20, 2009

### danago

One common method is to use the 'auxiliary angle method', which is to combine the sine and cosine terms into a single sine (or cosine) term of the form:

R sin (B+a).

The reason why we can write it in such a form is because this is equivalent to: (using an appropriate trig identity)

R sin(B)cos(a) + R cos(B)sin(a)

So by comparing coefficients, you can see that:

R cos a = 12
R sin a = 4

From which you can solve to find a and R. See if you can continue from there.

5. Mar 20, 2009

### Dell

so we have
R cos a = 4
R sin a = 12

tan a = 3
a=71.565
R=12.65

am i meant to be saying that

R sin (B+a) = R sin(B)cos(a) + R cos(B)sin(a) ??????????

therefore

R sin (B+a) = 1+ 2*[3*sin(39) - cos(39)]
sin(B+a)=0.25467

sin(c)=0.25467
c=14.75

c=B+71.565=14.75
B= (-56.81)

but this doesnt come right

cos(-56.81) + 3sin(-56.81)= -1.963
where i need
cosB + 3sinB=0.8054076

6. Mar 20, 2009

### danago

Yea the reason is because i made a mistake in my post but have since edited it out, and i think you have carried through that same mistake.

Comparing the two forms:

R sin(B)cos(a) + R cos(B)sin(a) = 4cos(B) + 12sin(B)

On the left hand side, the coefficient of cos(B) is R sin(a), and on the right hand side it is 4, and so we must have:

R sin(a) = 4

And by similar reasoning,:

R cos(a) = 12

So i had originally typed the 4 and 12 in the wrong places. I guess you read my post before i got a chance to edit it out (or perhaps coincidentally made the exact same mistake haha? :P)

7. Mar 20, 2009

### Dell

so...
R sin(a) = 4
R cos(a) = 12

tan(a)=1/3
(a)=18.34
R=12.65

R sin (B+a) = 1+ 2*[3*sin(39) - cos(39)]
sin(B+a)=0.25467

sin(c)=0.25467
c=14.75

c=B+18.34=14.75
B= (-3.59)

cos(-3.59) + 3sin(-3.59)= -0.81
thanks!!