# Thread: Sum and difference identities

1. ## Sum and difference identities part 2

Question cos 165
can it be solved by using the special angle
cos 150 + cos 15
cos 150 + cos (60-45)

Can it be solved in this way and will it provide me the correct answer

2. Originally Posted by mj.alawami
Angles for sum and difference identities?

how to find the angle of the sum and difference identities
for example sin 15 =(60-45) form where did the 60 and 45 come from and how did you get them in order to solve the problem
Why can't we use instead of 60 -45 >>>>>>10 +5?
Us the formula:

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

and see my attachment to see why we use 60 and 45 instead of 10 and 5. We can easily derive triangles that give us SOH CAH TOA expression for sin(30), sin(60), sin(45) and cos(30), cos(60) and cos(45).

We can not do the same for the angles 10 and 5.

3. Originally Posted by Mush
Us the formula:

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

and see my attachment to see why we use 60 and 45 instead of 10 and 5. We can easily derive triangles that give us SOH CAH TOA expression for sin(30), sin(60), sin(45) and cos(30), cos(60) and cos(45).

We can not do the same for the angles 10 and 5.
Can you solve my newly edited question please
Thank you very much for taking the time to help me

4. Originally Posted by mj.alawami
Question cos 165
can it be solved by using the special angle
cos 150 + cos 15
cos 150 + cos (60-45)

Can it be solved in this way and will it provide me the correct answer
Hrm not quite.

$\cos(165) = \cos(150+15) = \cos(150)\cos(15)-\sin(150)\sin(15)$

$= \cos(90 + 60)\cos(60-45) - \sin(90+60) \sin(60-45)$

$= \big(\cos(90)\cos(60)-\sin(90)\sin(60)\big)\big(\cos(60)\cos(45)+\sin(60 )\sin(45)\big) -$ $\big(\sin(90)\cos(60) + \sin(60)\cos(90)\big)\big(\sin(60)\cos(45)-\sin(45)\cos(60)\big)$

And you know that:

$\cos(90) = 0$

$\sin(90) = 1$

$\cos(60) = \frac{1}{2}$

$\sin(60) = \frac{\sqrt{3}}{2}$

$\cos(30) = \frac{\sqrt{3}}{2}$

$\sin(30) = \frac{1}{2}$

$\cos(45) = \frac{1}{\sqrt{2}}$

$\sin(45) = \frac{1}{\sqrt{2}}$

5. Originally Posted by Mush
Hrm not quite.

$\cos(165) = \cos(150+15) = \cos(150)\cos(15)-\sin(150)\sin(15)$

$= \cos(90 + 60)\cos(60-45) - \sin(90+60) \sin(60-45)$

$= \big(\cos(90)\cos(60)-\sin(90)\sin(60)\big)\big(\cos(60)\cos(45)+\sin(60 )\sin(45)\big) -$ $\big(\sin(90)\cos(60) + \sin(60)\cos(90)\big)\big(\sin(60)\cos(45)-\sin(45)\cos(60)\big)$

And you know that:

$\cos(90) = 0$

$\sin(90) = 1$

$\cos(60) = \frac{1}{2}$

$\sin(60) = \frac{\sqrt{3}}{2}$

$\cos(30) = \frac{\sqrt{3}}{2}$

$\sin(30) = \frac{1}{2}$

$\cos(45) = \frac{1}{\sqrt{2}}$

$\sin(45) = \frac{1}{\sqrt{2}}$
Can you use the diagram that you gave me to solve this problem please so it can be extremely clear ?
Thank you and sorry if i ask too much

6. Hello, mj.alawami!

Question: cos 165

can it be solved by using the special angle
cos 150 + cos 15
cos 150 + cos (60-45)

Can it be solved in this way and will it provide me the correct answer? . . . . Yes!
It is simpler with only two angles, of course . . .

Try: . $\cos165^o \;=\;\cos(120^o + 45^o)$

. . . . . . . . . $= \;\cos120^o\cos45^o - \sin120^o\sin45^o$

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

. . . . . . . . . $= \;-\frac{\sqrt{2}}{4} - \frac{\sqrt{2}\sqrt{3}}{4}$

. . . . . . . . . $=\;-\frac{\sqrt{2}(1 + \sqrt{3})}{4}$