1. ## trig differentiation

Hey i have this equation;
a² + b² - 2abCos Ѳ = c² + d² - 2cdCos ø
and i need to show that dѲ/dø = cd sinø / ab sinѲ

im not quite sure how to do this
i think that the 1st one goes to 2a + 2b + 2absinѲ
and the 2nd one goes to 2c + 2d + 2cdsinø

but im stuck from there

2. Originally Posted by Jampop
Hey i have this equation;
a² + b² - 2abCos Ѳ = c² + d² - 2cdCos ø
and i need to show that dѲ/dø = cd sinø / ab sinѲ

im not quite sure how to do this
i think that the 1st one goes to 2a + 2b + 2absinѲ
and the 2nd one goes to 2c + 2d + 2cdsinø

but im stuck from there
This problem would have been better off in the calculus section. You're not in a precalculus class are you? If so, this is a difficult problem.

Let $y=a^2+b^2-2abcos\theta$

Therefore we also have:

$y=c^2+d^2-2cdcos\phi$

$\frac{d\theta}{d\phi}=\frac{d\theta}{dy}\frac{dy}{ d\phi}$

Since $\frac{dy}{d\theta}=2absin\theta$, this means that $\frac{d\theta}{dy}=\frac{1}{2absin\theta}$.

$\frac{dy}{d\phi}=2cdcos\phi$

Substituting these derivatives into $\frac{d\theta}{d\phi}=\frac{d\theta}{dy}\frac{dy}{ d\phi}$ gives $\frac{d\theta}{d\phi}=\frac{cdsin\phi}{absin\theta }$

3. Oh ok i understand how you get it now but can you just help me with a few things?

Like how does

Why it is not 2a + 2b + 2absinѲ??

4. Originally Posted by Jampop
Oh ok i understand how you get it now but can you just help me with a few things?

Like how does

Why it is not 2a + 2b + 2absinѲ??
The numbers $a^2$ and $b^2$ are just constants. The derivative of a constant is zero. $\frac{d}{d\theta}a^2=0$. The same is true for $b^2$. Remember that a derivative is a rate of change with respect to some variable. In this case the variable is $\theta$. So the only part of the equation $a^2+b^2-2abcos\theta$ that changes with respect to $\theta$ is $-2abcos\theta$. The constants don't change (by definition of a constant), so their derivatives are zero.

5. Oh yeah thanks very muchh