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

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- Jun 8th 2008, 03:34 PMmissjkyangles 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 PMMathstud28
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 PMmasters
- Jun 8th 2008, 04:06 PMmissjky
[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 PMMathstud28
- Jun 8th 2008, 04:09 PMmissjky
- Jun 8th 2008, 04:25 PMo_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 PMmissjky
- Jun 8th 2008, 04:33 PMmissjky
- Jun 8th 2008, 04:36 PMo_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 PMmissjky
- Jun 8th 2008, 04:47 PMo_O
(Yes)