How do you prove #[sin(x+ y) - sin(x-y)] /[ cos(x+ y) + cos(x-y)]= tan y#?

Question

Asher Friedman · Accepted Answer

see explanation

To #color(blue)"Prove"# we require to manipulate one side into the same form as the other side.This will involve using #color(blue)"Addition formulae"# #color(orange)"Reminders"# #color(red)(|bar(ul(color(white)(a/a)color(black)(sin(A±B)=sinAcosB±cosAsinB)color(white)(a/a)|)))# #color(red)(|bar(ul(color(white)(a/a)color(black)(cos(A±B)=cosAcosB∓sinAsinB)color(white)(a/a)|)))# Starting with the left side and simplifying numerator/denominator separately. Numerator #sinxcosy+cosxsiny-[sinxcosy-cosxsiny)# #=cancel(sinxcosy)+cosxsiny-cancel(sinxcosy)+cosxsiny# #=2cosxsiny# Denominator #cosxcosy-sinxsiny+cosxcosy+sinxsiny# #=cosxcosy-cancel(sinxsiny)+cosxcosy+cancel(sinxsiny)# #=2cosxcosy# #"---------------------------------------------------------------"# left side can now be expressed as #(2cosxsiny)/(2cosxcosy)=(cancel(2)cancel(cosx)siny)/(cancel(2)cancel(cosx)cosy)=(siny)/(cosy)# and #(siny)/(cosy)=tany="right side hence proved"#

Emery Adkins · Answer

undefined To prove $ \frac{\sin(x+ y) - \sin(x-y)}{\cos(x+ y) + \cos(x-y)} = \tan y $, we'll use trigonometric identities.

1. Use the sum-to-product identities:
$$ \sin(a + b) = \sin a \cos b + \cos a \sin b $$
$$ \sin(a - b) = \sin a \cos b - \cos a \sin b $$
$$ \cos(a + b) = \cos a \cos b - \sin a \sin b $$
$$ \cos(a - b) = \cos a \cos b + \sin a \sin b $$

2. Substitute these identities into the expression:
$$ \frac{(\sin x \cos y + \cos x \sin y) - (\sin x \cos y - \cos x \sin y)}{(\cos x \cos y - \sin x \sin y) + (\cos x \cos y + \sin x \sin y)} $$

3. Simplify the numerator and denominator:
$$ \frac{\sin x \cos y + \cos x \sin y - \sin x \cos y + \cos x \sin y}{\cos x \cos y - \sin x \sin y + \cos x \cos y + \sin x \sin y} $$

4. Cancel out terms:
$$ \frac{2 \cos x \sin y}{2 \cos x \cos y} $$

5. Simplify further:
$$ \frac{\sin y}{\cos y} $$

6. Recall that $ \frac{\sin y}{\cos y} = \tan y $.

Hence, $ \frac{\sin(x+ y) - \sin(x-y)}{\cos(x+ y) + \cos(x-y)} = \tan y $ is proven.