I am kinda lost on this guys. The question is:
Find the exact value of the expression
Cos(300°+45°)
I am prty sure u have to use the angle formula, and i plugged it in, but i am lost now. anyone help me?
I am kinda lost on this guys. The question is:
Find the exact value of the expression
Cos(300°+45°)
I am prty sure u have to use the angle formula, and i plugged it in, but i am lost now. anyone help me?
300 = (270+30)
so we have cos((270+30)+45) = cos(270+30)cos(45) - sin(270+30)sin(45)
$\displaystyle cos(45) = sin(45) = \frac{1}{\sqrt2}$
$\displaystyle \frac{cos(270+30)-sin(270+30)}{\sqrt2}$
using cos(A+B) and sin(A+B) with A=270 and B=30 gives:
$\displaystyle \frac{[cos(270)cos(30) - sin(270)sin(30)] - [sin(270)cos(30) + cos(270)sin(30]}{\sqrt2}$
$\displaystyle cos(270) = 0$
$\displaystyle sin(270) = -1$
$\displaystyle cos (30) = \frac{1}{2}$
$\displaystyle sin(30) = \frac{\sqrt3}{2}$
$\displaystyle
\frac{[0+\frac{\sqrt3}{2}] - [\frac{\sqrt3}{4}+ 0]}{\sqrt2}$
$\displaystyle \frac{\sqrt3}{2} = \frac{2\sqrt3}{4}$:
$\displaystyle \frac{2\sqrt3 - \sqrt3}{4\sqrt2} = \frac{\sqrt3}{4\sqrt2} = \frac{\sqrt6}{8}$
perhaps there is a shorter way?
OK, clearly this is $\displaystyle \cos{345^\circ}$.
We could write this as $\displaystyle \cos{(360^\circ - 15^\circ)}$. So the angle is $\displaystyle 15^\circ$, just in the 4th quadrant. Remembering that cosine is positive in the fourth quadrant, we get
$\displaystyle \cos{(360^\circ - 15^\circ)} = \cos{15^\circ}$
This can be evaluated using a half-angle formula.
Remember that $\displaystyle \cos{\frac{\theta}{2}} = \sqrt{\frac{1 + \cos{\theta}}{2}}$.
So $\displaystyle \cos{15^\circ} = \cos{\frac{30^\circ}{2}}$
$\displaystyle = \sqrt{\frac{1 + \cos{30^\circ}}{2}}$
$\displaystyle = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}}$
$\displaystyle = \sqrt{\frac{1}{2} + \frac{\sqrt{3}}{4}}$.
So $\displaystyle \cos{(300^\circ + 45^\circ)} = \sqrt{\frac{1}{2} + \frac{\sqrt{3}}{4}}$.