Sum and Difference Identities

Math Analysis - Trigonometry - Sum and Difference Formulas - Sine - Cosine

Tip: This isn't the place to ask a question because the teacher can't reply.

1 of 3 videos by AJ Speller

Key Questions

• There are 2 identities you must know to simplify your expressions and solve equations

• sum identities :
${\left(a + b\right)}^{2} = {a}^{2} + {b}^{2} + 2 \cdot a \cdot b$
• difference identities:
${\left(a - b\right)}^{2} = {a}^{2} + {b}^{2} - 2 \cdot a \cdot b$

Proof:
${\left(a + b\right)}^{2} = \left(a + b\right) \cdot \left(a + b\right)$ then you expand
${\left(a + b\right)}^{2} = {a}^{2} + {b}^{2} + a \cdot b + b \cdot a$ and then $a \cdot b = b \cdot a$

${\left(a - b\right)}^{2} = \left(a - b\right) \cdot \left(a - b\right)$ then you expand
${\left(a - b\right)}^{2} = {a}^{2} + {b}^{2} - a \cdot b - b \cdot a$

Example:
${\left(x + 3\right)}^{2} = {x}^{2} + 9 + 6 x$
${\left(x - 3\right)}^{2} = {x}^{2} + 9 - 6 x$

Bonus:
$\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$
proof:
$\left(a + b\right) \left(a - b\right) = {a}^{2} - a \cdot b + b \cdot a - {b}^{2} = {a}^{2} - {b}^{2}$

• Here is an example:
sin(75) = sin(30 + 45) = sin(30) cos(45) + cos(30) sin(45)
= (1/2) ( $\sqrt{2}$/2) + ( $\sqrt{3}$/2)( $\sqrt{2}$/2)
= ( $\sqrt{2}$ + $\sqrt{6}$)/4

cos(75) = cos(30 + 45) = cos(30) cos(45) - sin(30) sin(45)
= ( $\sqrt{3}$/2)( $\sqrt{2}$/2) - (1/2) ( $\sqrt{2}$/2)
= ( $\sqrt{6}$ - $\sqrt{2}$)/4

sin(15) = sin(45 - 30) = sin(45) cos(30) - cos(45) sin(30)
= ( $\sqrt{2}$/2) ( $\sqrt{3}$/2) - ( $\sqrt{2}$/2)(1/2)
= ( $\sqrt{6}$ - $\sqrt{2}$)/4

cos(15) = cos(45 - 30) = cos(45) cos(30) + sin(45) sin(30)
=( $\sqrt{2}$/2) ( $\sqrt{3}$/2) + ( $\sqrt{2}$/2) (1/2)
=( $\sqrt{6}$ + $\sqrt{2}$)/4

• Main Sum and Differences Trigonometric Identities

$\cos \left(a - b\right) = \cos a \cdot \cos b + \sin a \cdot \sin b$
$\cos \left(a + b\right) = \cos a \cdot \cos b - \sin a \cdot \sin b$
$\sin \left(a - b\right) = \sin a \cdot \cos b - \sin b \cdot \cos a$
$\sin \left(a + b\right) = \sin a \cdot \cos b + \sin b \cdot \cos a$
$\tan \left(a - b\right) = \frac{\tan a - \tan b}{1 + \tan a \cdot \tan b}$
$\tan \left(a + b\right) = \frac{\tan a + \tan b}{1 - \tan a \cdot \tan b}$

Application of Sum and Differences Trigonometric Identities

Example 1: Find $\sin 2 a$.

$\sin 2 a$
$= \sin \left(a + a\right)$
$= \sin a \cdot \cos a + \sin a \cdot \cos a$
$= 2 \cdot \sin a \cdot \cos a$

Example 2: Find $\cos 2 a$.

$\cos 2 a$
$= \cos \left(a + a\right)$
$= \cos a \cdot \cos a - \sin a \cdot \sin a$
$= {\cos}^{2} a - {\sin}^{2} a$

Example 3: Find $\cos \left(\frac{13 \pi}{12}\right)$.

$\cos \left(\frac{13 \pi}{12}\right)$
$= \cos \left(\frac{\pi}{3} + \frac{3 \pi}{4}\right)$
$= \cos \left(\frac{\pi}{3}\right) \cdot \cos \left(\frac{3 \pi}{4}\right) - \sin \left(\frac{\pi}{3}\right) \cdot \sin \left(\frac{3 \pi}{4}\right)$
$= - \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4}$
$= - \frac{\sqrt{2} + \sqrt{6}}{4}$

• Here is an example of using a sum identity:

Find $\sin {15}^{\circ}$.

If we can find (think of) two angles $A$ and $B$ whose sum or whose difference is 15, and whose sine and cosine we know.

$\sin \left(A - B\right) = \sin A \cos B - \cos A \sin B$

We might notice that $75 - 60 = 15$
so $\sin {15}^{\circ} = \sin \left({75}^{\circ} - {60}^{\circ}\right) = \sin {75}^{\circ} \cos {60}^{\circ} - \cos {75}^{\circ} \sin {60}^{\circ}$

BUT we don't know sine and cosine of ${75}^{\circ}$. So this won't get us the answer. (I included it because when solving problems we DO sometimes think of approaches that won't work. And that's OK.)

$45 - 30 = 15$ and I do know the trig functions for ${45}^{\circ}$ and ${30}^{\circ}$

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

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

$= \frac{\sqrt{6} - \sqrt{2}}{4}$

There are other way of writing the answer.

Note 1
We could use the same two angles and the identity for $\cos \left(A - B\right)$ to find $\cos {15}^{\circ}$

Note 2
Instead of $45 - 30 = 15$ we could have used $60 - 45 = 15$

Note 3
Now that we have $\sin {15}^{\circ}$ we could use $60 + 15 = 75$ and $\sin \left(A + B\right)$ to find $\sin {75}^{\circ}$. Although if the question had been to find $\sin {75}^{\circ} , I ' d p r o b a b l y u s e$30^@$\mathmr{and}$45^@#

Questions

• · 4 weeks ago
• · 1 month ago
• · 2 months ago
• · 2 months ago
• · 2 months ago
• · 2 months ago
• · 2 months ago
• · 3 months ago
• · 3 months ago
• · 3 months ago
• · 3 months ago
• · 4 months ago
• · 4 months ago
• · 4 months ago
• · 5 months ago
• · 5 months ago
• · 5 months ago
• · 5 months ago