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
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
Us the formula:
$\displaystyle \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.
Hrm not quite.
$\displaystyle \cos(165) = \cos(150+15) = \cos(150)\cos(15)-\sin(150)\sin(15) $
$\displaystyle = \cos(90 + 60)\cos(60-45) - \sin(90+60) \sin(60-45) $
$\displaystyle = \big(\cos(90)\cos(60)-\sin(90)\sin(60)\big)\big(\cos(60)\cos(45)+\sin(60 )\sin(45)\big) -$$\displaystyle \big(\sin(90)\cos(60) + \sin(60)\cos(90)\big)\big(\sin(60)\cos(45)-\sin(45)\cos(60)\big) $
And you know that:
$\displaystyle \cos(90) = 0 $
$\displaystyle \sin(90) = 1 $
$\displaystyle \cos(60) = \frac{1}{2} $
$\displaystyle \sin(60) = \frac{\sqrt{3}}{2} $
$\displaystyle \cos(30) = \frac{\sqrt{3}}{2} $
$\displaystyle \sin(30) = \frac{1}{2} $
$\displaystyle \cos(45) = \frac{1}{\sqrt{2}} $
$\displaystyle \sin(45) = \frac{1}{\sqrt{2}} $
Hello, mj.alawami!
It is simpler with only two angles, of course . . .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!
Try: .$\displaystyle \cos165^o \;=\;\cos(120^o + 45^o)$
. . . . . . . . . $\displaystyle = \;\cos120^o\cos45^o - \sin120^o\sin45^o$
. . . . . . . . . $\displaystyle = \;\left(-\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right) - \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{2 }}{2}\right) $
. . . . . . . . . $\displaystyle = \;-\frac{\sqrt{2}}{4} - \frac{\sqrt{2}\sqrt{3}}{4}$
. . . . . . . . . $\displaystyle =\;-\frac{\sqrt{2}(1 + \sqrt{3})}{4}$