# solving trigonometry questions

• Nov 7th 2010, 01:21 AM
kelvinchy
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
• Nov 7th 2010, 06:57 AM
skeeter
Quote:

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 $\displaystyle \cos(x\pm y) = \cos{x}\cos{y} \mp \sin{x}\sin{y}$ for both the cosine sum and difference factors on the left side ... then simplify. then you'll need to rewrite the right side $\displaystyle \cos(3A)$ as $\displaystyle \cos(2A + A)$ and use the same identity to expand and simplify this expression. bottom line is ... lots of grunt work algebra in front of you.
$\displaystyle x = r\cos{\theta}$
$\displaystyle y = r\sin{\theta}$
$\displaystyle r = \sqrt{x^2+y^2}$
since $\displaystyle (x,y)$ is in quad II , $\displaystyle \theta = 180 + \arctan\left(\frac{y}{x}\right)$