Trigonometric expression manipulation

I'm working on manipulating trig expressions using Khan Academy.

According to the provided answer this:

$$\cos(\theta)\sin(\theta)$$

is supposed to equal:

$$\frac1 2\sin(2\theta)$$

I do not understand how, so could someone please show me.

• Do you wnt a proof that $\sin (2x)=2\sin x \cos x$ ? – Emilio Novati Nov 13 '15 at 13:51
• sin(2x) = sin(x+x) = ... use the addition formula. – jbuddenh Nov 13 '15 at 13:58

Hint: $\sin (x + y) =\sin (x) \cdot \cos (y) +\cos (x) \cdot \sin (y)$

What happens when $x=y$ ?

• Which begs the question: where does the first formula come from? – Simon S Nov 13 '15 at 15:57
• The practice session of manipulating trig expressions on KA comes after the exposure to the principal identities in question. – Nameless Nov 13 '15 at 16:56

Hint: Use the Law of Sines on the following diagram.

Hint 2: $\cos x = \sin(90^\circ - x)$

• I do not understand this – user252704 Nov 18 '15 at 11:13
• That's unfortunate, because it is reasonably straight forward. – John Joy Nov 18 '15 at 15:24

Hint:

It is an immediate consequence of the double angle formula $$\sin (2 \theta)= 2 \sin \theta \cos \theta$$

This formula is found, for $\alpha=\beta=\theta$ from the sum formula: $$\sin (\alpha+\beta)=\sin \alpha \cos \beta+ \cos\alpha \cos \beta$$and you can find a proof of this one here.

Using complex numbers, in particular $e^{i\theta} = cos(\theta) + isen(\theta)$

Using the change of variable: $\theta = 2\phi$

Whe have, $e^{i(2\phi)} = cos(2\phi) + isen(2\phi)$

And as $e^{i(2\phi)} = (e^{i\phi})^2 = (cos(\phi) + isen(\phi))^2 = cos^2(\phi) - sen^2(\phi) + i(2cos(\phi)sen(\phi))$

Then, $cos(2\phi) + isen(2\phi) = cos^2(\phi) - sen^2(\phi) + i(2cos(\phi)sen(\phi))$

Finally equalling real and imaginary parts:

$cos(2\phi) = cos^2(\phi) - sen^2(\phi)$

And

$sen(2\phi) = 2cos(\phi)sen(\phi)$

Which is what we where looking for.

$(e^{ia})^2 = e^{2ia}.$

L.H.S = $( cos^2(a) - sin^2(a) , 2sin(a).cos(a))$

R.H.S = $(cos(2a) , sin(2a))$