Question 826527
{{{sec(x) + sqrt(3)csc(x) = 4}}}
Looking at what we have and where we want to end up, {{{sin(x) + sqrt(3) cos( x) = 2 sin(2x)}}}, we should notice some things:<ul><li>There are no sec's or csc's at the end. So somehow we need to get rid of them.</li><li>Since sin(2x) = 2sin(x)cos(x), turning everything into sin's and cos's looks promising.</li></ul>Putting these together, along with the facts that a) sec and csc are reciprocals of cos and sin, respectively; and b) the product of reciprocals is always a 1, multiplying both sides by sin(x)cos(x) will so a lot of good:
{{{sin(x)cos(x)(sec(x) + sqrt(3)csc(x)) = sin(x)cos(x)(4)}}}
On the left side we need to use the Distributive Property:
{{{sin(x)cos(x)sec(x) + sin(x)cos(x)sqrt(3)csc(x) = sin(x)cos(x)(4)}}}
The reciprocal Trig functions will cancel each other out (since their product is 1):
{{{sin(x) + cos(x)sqrt(3) = 4sin(x)cos(x)}}}
If we use the Commutative Property on the second term, out left side is exactly what we wanted it to be:
{{{sin(x) + sqrt(3)cos(x) = 4sin(x)cos(x)}}}
Now we need to fix up the right side (without changing the left side). As we pointed out earlier, sin(2x) = 2sin(x)cos(x). If we factor a 2 out of the left side we get:
{{{sin(x) + sqrt(3)cos(x) = 2*2sin(x)cos(x)}}}
And we can now substitute in the sin(2x):
{{{sin(x) + sqrt(3)cos(x) = 2*sin(2x)}}}