# Trigonometry help!

If 3sinA+5cosA=5 show that 5sinA-3cosA=+-3

 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #2 May 11, 2011, 10:03 AM
$3\sin A + 5\cos A = 5$

You'll have to convert it to something like $\sin (A + \alpha)$ or $\cos(A+\alpha)$. Both are acceptable, but I personally find sin to be easier because it is positive for the longest interval when you start.

So, let: $3\sin A + 5\cos A \equiv R\sin(A + \alpha)$

Using the compound angle formula, you get the right side to be:

$R(\sin A\cos\alpha + \cos A\sin\alpha) = R\sin A\cos\alpha + R\cos A\sin\alpha$

Then you equate that with the left hand side:

$3\sin A = R\sin A\cos\alpha$

$5\cos A = R\cos A\sin\alpha$

Find R and alpha and then replace those values in your original equation.

$R\sin(A + \alpha) = 5$

This way, it's easier:

$A= \sin^{-1}$$\frac{5}{R}$$ - \alpha$

Now that you have A, the next part should be easy.
 jcaron2 Posts: 983, Reputation: 1034 Senior Member #3 May 11, 2011, 03:01 PM
Quote:
 Originally Posted by Unknown008 Now that you have A, the next part should be easy.
Easy if you have a calculator and can compute the inverse sine to get the angle. However, maybe not so easy if you try to solve it with a closed-form expression. Though I have to say, I like the way you're essentially deriving the equation for linear combination of sines and cosines by breaking the one equation into two separate ones (since sine and cosine are orthogonal functions). For what it's worth, if you just look up the formula, I think you get R=sqrt(34) (i.e. Sqrt(3^2+5^2)) and alpha=atan(3/5), which is exactly the same as what your method should give. Nice one, Jerry.

I skipped all that and just did it with phasor addition, which makes it almost trivial. The left side of the first equation results in a phasor of length sqrt(34) at an angle of atan(3/5). The left side of the second equation results in a phasor with the same length at an angle of atan(-5/3). Since the tangents are negative reciprocals of each other, the two phasors are clearly perpendicular. Rotating the phasors together (which indicates a changing angle A), there are only two places on the circle where the real part of the first phasor is 5. The first is when there is no rotation at all (the x-component of the phasor was already 5 to begin with). In that case, the x-component of the second phasor is -3. The second case is when the pair are rotated until the first phasor is at the negative of its original angle (an angle of pi+atan(-3/5)). That puts the second phasor at an angle of atan(5/3), so it's x-component is +3.

Much easier explained with a picture than with words. See the attachment. I drew the phasors for the first equation in black, the second equation in red. The nominal angle with solid lines and the rotated angle with dashed lines. Note that in the two cases (solid and dashed), the x-coordinates of the red phasors comes out to +/-3.
Attached Images

 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #4 May 12, 2011, 09:34 AM
Nice proof Josh! I never pictured it that way, really neat!