Trig Equation (1 Viewer)

Axio · Jul 20, 2014

Hey guys I'm having a bit of difficulty with this question:

Write the expression cosy - (sqroot of 3) x siny in the form: r cos(y + a)

Thanks

Drongoski · Jul 20, 2014

Axio said:
Hey guys I'm having a bit of difficulty with this question:

Write the expression cosy - (sqroot of 3) x siny in the form: r cos(y + a)

Thanks

Textbooks offer a well-known standard method (see faisal's method below). I'll do it slightly differently. Sketching a right angled-triangle with sides 1 and sqrt(3) and therefore hypotenuse =2 will help.

$LHS = 1 cos y - \sqrt 3 siny = 2[\frac{1}{2} cos y - \frac {\sqrt 3}{2} siny] = 2[cos \frac {\pi}{3} cos y - sin \frac {\pi}{3} siny] \\ \\ = 2 cos(y + \frac {\pi}{3})$

faisalabdul16 · Jul 20, 2014

Axio said:
Hey guys I'm having a bit of difficulty with this question:

Write the expression cosy - (sqroot of 3) x siny in the form: r cos(y + a)

Thanks

R = square root of (1^2 + (root3)^2) = 2. your angle is just tanA = root3. Therefore A = pi/3.

2cos(y+pi/3)

Axio · Jul 20, 2014

Drongoski said:
Textbooks offer a well-known standard method. I'll do it slightly differently. Sketching a right angled-triangle with sides 1 and sqrt(3) and therefore hypotenuse =2 will help.

$LHS = 1 cos y - \sqrt 3 siny = 2[\frac{1}{2} cos y - \frac {\sqrt 3}{2} siny] = 2[cos \frac {\pi}{3} cos y - sin \frac {\pi}{3} siny] \\ \\ = 2 cos(y + \frac {\pi}{3})$

Thanks, that's a good method

.

faisalabdul16 said:
R = square root of (1^2 + (root3)^2) = 2. your angle is just tanA = root3. Therefore A = pi/3.

2cos(y+pi/3)

Thanks

photastic · Jul 20, 2014

Tbh! I just accepted the transformation method because it is bloody ambigious. I just stuck to the generic formulas.

Axio · Jul 20, 2014

faisalabdul16 said:
A quicker way to getting the angle rather than doing the double angle thingy for cos, is just doing sin/cos = the coefficient of sin/ coefficient of cos. Its a quick trick way to getting the angle quickly. Im not sure if you understood what i just said, hard to explain aha

It'd be good if you did an example

.

aDimitri · Jul 20, 2014

Auxiliary angle method involves taking an expression in the form asinx + bcosx and changing it into a single function. This is the method that involves dividing to get tana in terms of a constant. in this case the question was

$cosy - \sqrt{3}siny \\ \text{We want it in the form} \\ Rcos(y+a) \\ \text{By expanding,} \\ Rcos(y+a) = Rcosacosy - Rsinasiny \\ \text{Comparing this with the original expression,} \\ Rcosa = 1 \: \: \: \: \text{ (A)} \\ Rsina = \sqrt{3} \: \: \: \: \text{ (B)}\\ (A)^{2} + (B)^2 \\ R^{2}(sin^{2}a + cos^{2}a) = 1^{2} + \sqrt{3}^{2} = 4 \\ R = \sqrt{4} = 2 \\ (B) \div (A) \\ tana = \frac{\sqrt{3}}{1} \\ a = \frac{\pi}{3} \\ \text{Therefore, } cosy - \sqrt{3}siny = 2cos\left(y+\frac{\pi}{3}\right)$

Hope this helps

Trig Equation (1 Viewer)

Axio

=o

Drongoski

Well-Known Member

faisalabdul16

Wot

Axio

=o

photastic

Well-Known Member

Axio

=o

aDimitri

i'm the cook

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