How do i solve this trig question? identities etc?

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Homework Help Overview

The discussion revolves around solving a trigonometric equation involving identities. The original poster presents an equation that combines cosine and sine functions and seeks to find the value of the variable B.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore methods to combine sine and cosine terms into a single trigonometric function. The auxiliary angle method is suggested as a potential approach. There are discussions about comparing coefficients and resolving errors in initial assumptions regarding the terms.

Discussion Status

Participants are actively engaging with the problem, correcting each other's mistakes, and refining their understanding of the relationships between the terms. There is an ongoing exploration of how to accurately apply the auxiliary angle method and resolve discrepancies in calculations.

Contextual Notes

There are indications of previous mistakes in the setup of equations, and participants are questioning the accuracy of their calculations and assumptions throughout the discussion.

Dell
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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
 
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Can you think of a way to perhaps combine the sine and cosine terms of the left into a single trig function?
 
no, that's what i need to somehow do
 
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.
 
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 doesn't come right

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

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