# angles and trigonometic functions

• Jun 8th 2008, 03:34 PM
missjky
angles and trigonometic functions
Find the exact value of sin 30 ° - sin 45 ° cos 45 ° i dnt unnderstand how u find the value
• Jun 8th 2008, 04:00 PM
Mathstud28
Quote:

Originally Posted by missjky
Find the exact value of sin 30 ° - sin 45 ° cos 45 ° i dnt unnderstand how u find the value

You have to use your knowledge of trig exact values and your use of trig identites

$\displaystyle \sin(\psi)\cos(\psi)=\frac{1}{2}\sin(2\psi)$

$\displaystyle \therefore\sin(45)\cos(45)=\frac{1}{2}\sin(90)=\fr ac{1}{2}$

and it is commonly known that $\displaystyle \sin(30)=\frac{1}{2}$

So we have

$\displaystyle \frac{1}{2}-\frac{1}{2}=0$
• Jun 8th 2008, 04:03 PM
masters
Quote:

Originally Posted by missjky
Find the exact value of sin 30 ° - sin 45 ° cos 45 ° i dnt unnderstand how u find the value

In a 30-60-90 deg. right triangle, sin 30 = 1/2

In a 45-45-90 deg. right triangle, sin 45 = cos 45 = $\displaystyle \frac{\sqrt2}{2}$

$\displaystyle \sin 30-\sin 45 \cos45 = \frac{1}{2}-(\frac{\sqrt2}{2})(\frac{\sqrt2}{2})=\frac{1}{2}-\frac{2}{4}=0$
• Jun 8th 2008, 04:06 PM
missjky
[quote=masters;155321]

thanks so much and also the exact value of cos http://ce.byu.edu/courses/hs/9990640.../TRIG41-52.gif
• Jun 8th 2008, 04:07 PM
Mathstud28
[quote=missjky;155322]
Quote:

Originally Posted by Mathstud28
thanks oso much i understand and also
the exact value of cos http://ce.byu.edu/courses/hs/9990640.../TRIG41-52.gif

$\displaystyle \cos\bigg(\frac{\pi}{6}\bigg)=\frac{\sqrt{3}}{2}$
• Jun 8th 2008, 04:09 PM
missjky
[quote=Mathstud28;155325]
Quote:

Originally Posted by missjky
$\displaystyle \cos\bigg(\frac{\pi}{6}\bigg)=\frac{\sqrt{3}}{2}$

If θ is an acute angle in standard position and cot θ = 12/5 how do u find sin θ
• Jun 8th 2008, 04:25 PM
o_O
$\displaystyle \cot \theta = \frac{12}{5}$

I don't know what is meant by 'standard position' but I'm assuming that the angle is restricted to Quadrant I.

If you recall the definition that $\displaystyle \cot \theta = \frac{\text{adjacent}}{\text{opposite}}$, then you can draw a right angled triangle with a base of 12 and a height of 5 with $\displaystyle \theta$ as your angle.

Recall that $\displaystyle \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$. We already have the 'opposite' side, i.e. 5. We just need the hypotenuse which you can easily calculate with Pythagoras' Theorem.
• Jun 8th 2008, 04:30 PM
missjky
Quote:

Originally Posted by o_O
$\displaystyle \cot \theta = \frac{12}{5}$

I don't know what is meant by 'standard position' but I'm assuming that the angle is restricted to Quadrant I.

If you recall the definition that $\displaystyle \cot \theta = \frac{\text{adjacent}}{\text{opposite}}$, then you can draw a right angled triangle with a base of 12 and a height of 5 with $\displaystyle \theta$ as your angle.

Recall that $\displaystyle \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$. We already have the 'opposite' side, i.e. 5. We just need the hypotenuse which you can easily calculate with Pythagoras' Theorem.

i dnt no how to do that this is my first time taking trig plus its online
• Jun 8th 2008, 04:33 PM
missjky
Quote:

Originally Posted by missjky
i dnt no how to do that this is my first time taking trig plus its online

the asnswr is either
12/13
5/13
13/12
13/5
• Jun 8th 2008, 04:36 PM
o_O
Do you know what $\displaystyle \cot \theta$ and $\displaystyle \sin \theta$ stand for in terms of a triangle? You should be able to figure it out.

Here's a diagram to help illustrate what I'm trying to say. Solve for h and you should be able to find $\displaystyle \sin \theta$
• Jun 8th 2008, 04:47 PM
missjky
Quote:

Originally Posted by o_O
Do you know what $\displaystyle \cot \theta$ and $\displaystyle \sin \theta$ stand for in terms of a triangle? You should be able to figure it out.

Here's a diagram to help illustrate what I'm trying to say. Solve for h and you should be able to find $\displaystyle \sin \theta$

would it b 5/13
• Jun 8th 2008, 04:47 PM
o_O
(Yes)