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
$\displaystyle \frac{1}{2}-\frac{1}{2}=0$
$\displaystyle \cot \theta = \frac{12}{5}$
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.
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 $\displaystyle \sin \theta$