# Sum or Difference Formulas

• Jan 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)

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.
• Jan 4th 2012, 10:43 PM
MarkFL
Re: Sum or Difference Formulas
1.) $\displaystyle \sin\left(165^{\circ}\right)$

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

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

Now apply the angle difference identity for sine.

Or alternately:

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

Now apply the angle sum identity for cosine.

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

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

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

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

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

Now apply the angle difference identity for sine.

Or, alternately:

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

Now apply the angle sum identity for cosine.
• Jan 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.

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

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

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

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

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

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

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

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

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

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

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

• Jan 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?

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

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

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

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

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

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

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

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

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

Unfortunately I do not know the formulas.

Thanks for the help though.
• Jan 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
• Jan 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...
• Jan 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.