$$ \sin (x±y)= \sin x\cos y±\cos x \sin y $$
$$ sin(2\theta)=2\sin \theta \cos \theta $$
Recalled the add and subtraction identities. Now, what
if x and y (u and v) are the same?
If we plug in the same value x into the identities of add identities of sine. The equation turns to this.
$\sin(x+x)\sin x \cos x+ \cos x \sin x$
What is x +x ? If there is x equal terms(u) been adds to together. Then the x number of terms(u) means x times the term u.
<aside> 💡 So $\sin(2\theta)-\sin(\theta)$ Cannot be factor by sin of theta. Argument should be the same.
</aside>
$$ \cos (x±y) = \cos x\cos y ∓\sin x\sin y $$
$$ \cos(2\theta)=2\cos^2\theta -1 $$
$$ \cos(2\theta)=1-2\sin^2\theta $$
Recalled the add and subtraction identities. Now, what
if x and y (u and v) are the same?
If we plug in the same value x into the identities of add identities of cosine. The equation turns to this.
$\cos(x+x)=\cos x \cos x- \sin x \sin x$
to this $\cos(x+x)=\cos^2 x- \sin^2 x$
Remember the Pythagorean Identities? From there, we branches two version of cosine’s double angle identities
$\cos (x+x)=\cos x^2-(1-\cos^2x)$
final
$$ \cos(2\theta)=2\cos^2\theta -1 $$
$\cos(x+x)=(1-\sin^2x)-\sin^2x$
final
$$ \cos(2\theta)=1-2\sin^2\theta $$
$$ \tan (x±y) =\frac {\tan x±\tan y} {1∓\tan x\tan y} $$
$$ \tan(2\theta )=\frac {2\tan \theta } {1- \tan ^2\theta} $$
Recalled the add and subtraction identities. Now, what
if x and y (u and v) are the same?
If we plug in the same value x into the identities of add identities of sine. The equation turns to this.
$$ \tan(x+x)=\frac {\tan x +\tan x} {1-\tan x\tan x} $$
Remember to simplify the top and the bottom.