SOLUTION: Show that the equation sec x + sqrt3 cosec x = 4 can be written in the form sin x + sqrt3 cos x = 2 sin 2x

Algebra ->  Trigonometry-basics -> SOLUTION: Show that the equation sec x + sqrt3 cosec x = 4 can be written in the form sin x + sqrt3 cos x = 2 sin 2x      Log On


   

Question 826527: Show that the equation sec x + sqrt3 cosec x = 4 can be written in the form
sin x + sqrt3 cos x = 2 sin 2x
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
sec%28x%29+%2B+sqrt%283%29csc%28x%29+=+4
Looking at what we have and where we want to end up, sin%28x%29+%2B+sqrt%283%29+cos%28+x%29+=+2+sin%282x%29, we should notice some things:
  • There are no sec's or csc's at the end. So somehow we need to get rid of them.
  • Since sin(2x) = 2sin(x)cos(x), turning everything into sin's and cos's looks promising.
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%28x%29cos%28x%29%28sec%28x%29+%2B+sqrt%283%29csc%28x%29%29+=+sin%28x%29cos%28x%29%284%29
On the left side we need to use the Distributive Property:

The reciprocal Trig functions will cancel each other out (since their product is 1):
sin%28x%29+%2B+cos%28x%29sqrt%283%29+=+4sin%28x%29cos%28x%29
If we use the Commutative Property on the second term, out left side is exactly what we wanted it to be:
sin%28x%29+%2B+sqrt%283%29cos%28x%29+=+4sin%28x%29cos%28x%29
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%28x%29+%2B+sqrt%283%29cos%28x%29+=+2%2A2sin%28x%29cos%28x%29
And we can now substitute in the sin(2x):
sin%28x%29+%2B+sqrt%283%29cos%28x%29+=+2%2Asin%282x%29