1. ## solving trigonometry questions

1.
prove that cosAcos(30-A)cos(30+A)= 1/4 cos3A
Hence, evaluate cosAcos(30-A)cos(30+A) when A= 170 degree;

2.
Assuming r is positive, find r and the value of θ between -180degree and 180 degree which satisfy the questions:
r cos θ= -6 and r sin θ= 1.5

2. Originally Posted by kelvinchy
1.
prove that cosAcos(30-A)cos(30+A)= 1/4 cos3A
Hence, evaluate cosAcos(30-A)cos(30+A) when A= 170 degree;

2.
Assuming r is positive, find r and the value of θ between -180degree and 180 degree which satisfy the questions:
r cos θ= -6 and r sin θ= 1.5
for#1, you'll need to use the cosine sum/difference identity $\cos(x\pm y) = \cos{x}\cos{y} \mp \sin{x}\sin{y}$ for both the cosine sum and difference factors on the left side ... then simplify. then you'll need to rewrite the right side $\cos(3A)$ as $\cos(2A + A)$ and use the same identity to expand and simplify this expression. bottom line is ... lots of grunt work algebra in front of you.

for #2, note the polar/cartesian equations below ...

$x = r\cos{\theta}
$

$y = r\sin{\theta}$

$r = \sqrt{x^2+y^2}$

since $(x,y)$ is in quad II , $\theta = 180 + \arctan\left(\frac{y}{x}\right)$