0
$\begingroup$

$$2\sin(45^\circ-x)\cos(45^\circ-x)$$

I know you have to use the double angle formula for sine, but what next?

$\endgroup$
1
  • $\begingroup$ Yeah, I got up to that, but the answer states cos 2x. How do you expand sin 2(45-x)? $\endgroup$ Commented Apr 8, 2013 at 8:33

3 Answers 3

3
$\begingroup$

Using the double angle formula for sine, $$2\sin(\theta)\cos(\theta)=\sin(2\theta),$$ we see that $$2\sin(45-x)\cos(45-x)=\sin(2\cdot(45-x))=\sin(90-2x)$$ Now use the identity that allows you to simplify $\sin(90-\theta)$, where in our case we have $\theta=2x$. You'll then want to use the double angle formula for cosine.

$\endgroup$
2
  • $\begingroup$ Thanks! That cleared things up! $\endgroup$ Commented Apr 8, 2013 at 8:37
  • 1
    $\begingroup$ Glad I could help! $\endgroup$ Commented Apr 8, 2013 at 8:37
3
$\begingroup$

Hint : use $$2 \sin a \cdot \cos a = \sin 2a$$ and $$\sin(90-a)=\cos a$$

$\endgroup$
1
$\begingroup$

double angle formula for sine,

$2sin(\theta)cos(\theta)=sin(2\theta)$,

now,

$2sin(45−x)cos(45−x)=sin(2⋅(45−x))=sin(90−2x)$.

$sin(90-2x)=cos2x$

since,$(90-\theta)=cos(\theta)$, since cosine is positive in first and fourth quadrant, and maximum value of $\theta$ can not exceed $180^o$.

Now, Its easy ahead

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .