PRECALCULUS (Trig Functions) -SO LOST

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November 30th 2006, 04:11 PM
Whitewolfblue

PRECALCULUS (Trig Functions) -SO LOST

I am having a serous issue with these two problems. No matter what formula's I use i keep getting the wrong answer. I NEED HELP BADLY.

1.) Use a sum or difference formula or a half angle formula to determine the value of the trigonometric functions. Give exact answers. Do not use decimal numbers. The answer should be a fraction or an arithmetic expression. If the answer involves a square root it should be enter as sqrt; e.g. the square root of 2 should be written as sqrt(2);

sin((11pi)/8) =
sin(-(pi/12)) =
cos((3pi)/8) =
cos((11pi)/8) =

&

2.) If cos(t) = - (3/5) where pi < t < ((3pi)/2), find the values of the following trigonometric

Give exact answers, do not use decimal numbers. The answer should be a fraction or an arithmetic expression. If the answer involves a square root it should be enter as sqrt; e.g. the square root of 2 should be written as
sqrt(2).

cos(2t) =
sin(2t) =
cos(t/2) =
sin(t/2) =

Any and all Help greatly appreciated

~Kiba~
November 30th 2006, 05:49 PM
topsquark

Quote:

Originally Posted by Whitewolfblue

2.) If cos(t) = - (3/5) where pi < t < ((3pi)/2), find the values of the following trigonometric

Give exact answers, do not use decimal numbers. The answer should be a fraction or an arithmetic expression. If the answer involves a square root it should be enter as sqrt; e.g. the square root of 2 should be written as
sqrt(2).

cos(2t) =
sin(2t) =
cos(t/2) =
sin(t/2) =

$cos(t) = -\frac{3}{5}$

so
$sin(t) = \pm \sqrt{1 - cos^2(t)} = \pm \sqrt{ \frac{16}{25} } = \pm \frac{4}{5}$
Since $\pi < t < \frac{3 \pi}{2}$ we know to use the "-" sign. So
$sin(t) = - \frac{4}{5}$

Now.
$cos(2t) = 2cos^2(t) - 1 = 2 \frac{9}{25} - 1 = \frac{18 - 25}{25} = -\frac{7}{25}$

$sin(2t) = 2 sin(t) cos(t) = 2 \cdot \frac{-4}{5} \cdot \frac{-3}{5} = \frac{24}{25}$

$cos(t/2) = \pm \sqrt{ \frac{1 + cos(t)}{2} } = \pm \sqrt{ \frac{1 - \frac{3}{5}}{2} } = \pm \sqrt{\frac{1}{5}}$

Since $\pi < t < \frac{3 \pi}{2}$ we know that $\frac{ \pi}{2} < \frac{t}{2} < \frac{3 \pi}{4} < \pi$ so t/2 is in the second quadrant. This means we use the "-" sign.

Thus
$cos(t/2) = - \frac{1}{\sqrt{5}}$

$sin(t/2) = \pm \sqrt{ \frac{1 - cos(t)}{2} } = \pm \sqrt{ \frac{1 - -\frac{3}{5}}{2} } = \pm \sqrt{ \frac{4}{5} } = \pm \frac{2}{\sqrt{5}}$

Since t/2 is in the second quadrant, we use the "+" sign, so
$sin(t/2) = \frac{2}{\sqrt{5}}$

-Dan
November 30th 2006, 06:09 PM
topsquark

Quote:

Originally Posted by Whitewolfblue

1.) Use a sum or difference formula or a half angle formula to determine the value of the trigonometric functions. Give exact answers. Do not use decimal numbers. The answer should be a fraction or an arithmetic expression. If the answer involves a square root it should be enter as sqrt; e.g. the square root of 2 should be written as sqrt(2);

sin((11pi)/8) =
sin(-(pi/12)) =
cos((3pi)/8) =
cos((11pi)/8) =

$sin(11\pi/8) = sin(\pi + 3\pi/8) =$ $sin(\pi)cos(3 \pi/8) + sin(3\pi/8)cos(\pi) = -sin(3\pi/8)$

= $-sin \left ( \frac{1}{2} \cdot \frac{3 \pi}{4} \right )$

= $- \sqrt{ \frac{1 - cos(3 \pi/4)}{2} } = - \sqrt{ \frac{1 + cos(\pi/4)}{2} }$ (Using the reference angle)

= $- \sqrt{ \frac{1 + \frac{\sqrt{2}}{2}}{2} } = - \sqrt{\frac{2 + \sqrt{2}}{4}}$
================================================== =====

$cos(11\pi/8) = cos(\pi + 3\pi/8) =$ $cos(\pi) cos(3\pi/8) - sin(\pi) sin(3\pi/8) = - cos(3\pi/8)$

= $-cos \left ( \frac{1}{2} \cdot \frac{3 \pi}{4} \right )$

= $- \sqrt{ \frac{1 + cos(3 \pi/4)}{2} } = -\sqrt{ \frac{1 - cos(\pi/4)}{2} }$

= $- \sqrt{ \frac{1 - \frac{\sqrt{2}}{2}}{2} } = - \sqrt{ \frac{2 - \sqrt{2}}{4}}$
================================================== =====

$sin(-\pi/12) = -sin \left ( \frac{1}{2} \cdot \frac{\pi}{6} \right )$

= $- \sqrt{ \frac{1 - cos(\pi/6)}{2} } = -\sqrt{ \frac{1 - \frac{\sqrt{3}}{2}}{2} }$

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

And I already showed you what to do with $cos(3 \pi/8)$ in the second problem.

-Dan