Sum or Difference Formulas

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January 4th 2012, 10:27 PM
Nickod777

Sum or Difference Formulas

This is not for homework, nor for a teacher to see, just a test correction so that I know what to do, or what I am doing wrong.

Use the sum or Difference formulas to find the exact value of each expression.

1. sin (165 degrees)

2. cos (5π/12)

I tried to use my calculator, a TI-89 ended up with the same answer both, and it was wrong unfortunately, and I was possibly wondering what I could do to get the correct answer.

my TI-89 gave me this:

((√3-1)*√(2))/(4)
Radical 3 - 1 * radical 2 all over 4.

And my best guess as the answer for both is [√6 + √2] / 4. Just not sure how to get that.

Edit: this might be my only trig question on here, as the 1st semester is almost gone, and I just wanted to figure these problems out.
January 4th 2012, 10:43 PM
MarkFL

Re: Sum or Difference Formulas

1.) $\sin\left(165^{\circ}\right)$

Use the reflection $\sin\left(180^{\circ}-x\right)=\sin(x)$

$\sin\left((180-165)^{\circ}\right)=\sin\left$(45-30)^{\circ}$$

Now apply the angle difference identity for sine.

Or alternately:

$\sin\left(165^{\circ} \right)=\sin\left(15^{\circ}\right)=\cos\left(75^{ \circ}\right)=\cos\left((45+30)^{\circ}\right)$

Now apply the angle sum identity for cosine.

2.) $\cos\left(\frac{5\pi}{12}\right)$

Apply the identity: $\cos(x)=\sin\left(\frac{\pi}{2}-x\right)$ to get:

$\cos\left(\frac{5\pi}{12} \right)=\sin\left(\frac{\pi}{2}-\frac{5\pi}{12}\right)$

$\cos\left(\frac{5\pi}{12} \right)=\sin\left(\frac{\pi}{12}\right)$

$\cos\left(\frac{5\pi}{12} \right)=\sin\left(\frac{\pi}{4}-\frac{\pi}{6}\right)$

Now apply the angle difference identity for sine.

Or, alternately:

$\cos\left(\frac{5\pi}{12}\right)=\cos\left(\frac{( 3+2)\pi}{12}\right)=\cos\left(\frac{\pi}{4}+\frac{ \pi}{6}\right)$

Now apply the angle sum identity for cosine.
January 5th 2012, 05:44 AM
Soroban

Re: Sum or Difference Formulas

Hello, Nickod777!

Quote:

Use the Sum or Difference formulas to find the exact value of each expression.

. . $1.\;\sin (165^o) \qquad 2.\;\cos\left(\tfrac{5\pi}{12}\right)$

Did it occur to you to use the Sum or Difference Formulas?
Do you know the Sum or Difference Formulas?

$1.\;\sin(165^o) \;=\;\sin(45^o + 120^o)$

. . . . . . . . . $=\; \sin(45^o)\cos(120^o) + \cos(45^o)\sin(120^o)$

. . . . . . . . . $=\;\left(\frac{1}{\sqrt{2}}\right)\left(-\frac{1}{2}\right) + \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3 }}{2}\right)$

. . . . . . . . . $=\; -\frac{1}{2\sqrt{2}} + \frac{\sqrt{3}}{2\sqrt{2}} \;=\;\frac{\sqrt{3}-1}{2\sqrt{2}} \quad\begin{array}{c}^{\text{Usually this is an acceptable answer,}} \\ ^{\text{but often they rationalize the denominator.}}\end{array}$

. . . . . . . . . $=\;\frac{\sqrt{2}}{\sqrt{2}}\cdot \frac{\sqrt{3}-1}{2\sqrt{2}} \;=\; \frac{\sqrt{2}\left(\sqrt{3}-1\right)}{4} \;=\;\frac{\sqrt{6} - \sqrt{2}}{4}$

$2.\;\cos\left(\frac{5\pi}{12}\right) \;=\;\cos\left(\frac{\pi}{6} + \frac{\pi}{4}\right)$

. . . . . . . . . $=\;\cos\left(\frac{\pi}{6}\right)\cos\left(\frac{\ \pi}{4}\right) - \sin\left(\frac{\pi}{6}\right)\sin\left(\frac{\pi} {4}\right)$

. . . . . . . . . $=\;\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{\ \sqrt{2}}\right) - \left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{2}} \right)$

. . . . . . . . . $=\;\frac{\sqrt{3}-1}{2\sqrt{2}}$
January 5th 2012, 04:29 PM
Nickod777

Re: Sum or Difference Formulas

Quote:

Originally Posted by Soroban

Hello, Nickod777!

Did it occur to you to use the Sum or Difference Formulas?
Do you know the Sum or Difference Formulas?

$1.\;\sin(165^o) \;=\;\sin(45^o + 120^o)$

. . . . . . . . . $=\; \sin(45^o)\cos(120^o) + \cos(45^o)\sin(120^o)$

. . . . . . . . . $=\;\left(\frac{1}{\sqrt{2}}\right)\left(-\frac{1}{2}\right) + \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3 }}{2}\right)$

. . . . . . . . . $=\; -\frac{1}{2\sqrt{2}} + \frac{\sqrt{3}}{2\sqrt{2}} \;=\;\frac{\sqrt{3}-1}{2\sqrt{2}} \quad\begin{array}{c}^{\text{Usually this is an acceptable answer,}} \\ ^{\text{but often they rationalize the denominator.}}\end{array}$

. . . . . . . . . $=\;\frac{\sqrt{2}}{\sqrt{2}}\cdot \frac{\sqrt{3}-1}{2\sqrt{2}} \;=\; \frac{\sqrt{2}\left(\sqrt{3}-1\right)}{4} \;=\;\frac{\sqrt{6} - \sqrt{2}}{4}$

$2.\;\cos\left(\frac{5\pi}{12}\right) \;=\;\cos\left(\frac{\pi}{6} + \frac{\pi}{4}\right)$

. . . . . . . . . $=\;\cos\left(\frac{\pi}{6}\right)\cos\left(\frac{\ \pi}{4}\right) - \sin\left(\frac{\pi}{6}\right)\sin\left(\frac{\pi} {4}\right)$

. . . . . . . . . $=\;\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{\ \sqrt{2}}\right) - \left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{2}} \right)$

. . . . . . . . . $=\;\frac{\sqrt{3}-1}{2\sqrt{2}}$

Unfortunately I do not know the formulas.

Thanks for the help though.
January 5th 2012, 06:01 PM
pickslides

Re: Sum or Difference Formulas

Quote:

Originally Posted by Nickod777

Unfortunately I do not know the formulas.

Thanks for the help though.

Here are the formulas MarkFL2 & Soroban used.

Trigonometric Formulas: Sum and Difference
January 5th 2012, 06:07 PM
Prove It

Re: Sum or Difference Formulas

Quote:

Originally Posted by Nickod777

Unfortunately I do not know the formulas.

I find it hard to believe that you were asked to use formulae if you haven't been taught them...
January 5th 2012, 06:53 PM
Nickod777

Re: Sum or Difference Formulas

Quote:

Originally Posted by Prove It

I find it hard to believe that you were asked to use formulae if you haven't been taught them...

in my class, we are given the formulas, not learning how to use them, instead we have to learn them ourselves.

Anyway, let us just end this gentlemen.