document.write( "Question 826527: Show that the equation sec x + sqrt3 cosec x = 4 can be written in the form
\n" ); document.write( "sin x + sqrt3 cos x = 2 sin 2x
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Algebra.Com's Answer #498147 by jsmallt9(3758)\"\" \"About 
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\"sec%28x%29+%2B+sqrt%283%29csc%28x%29+=+4\"
\n" ); document.write( "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:
\n" ); document.write( "\"sin%28x%29cos%28x%29%28sec%28x%29+%2B+sqrt%283%29csc%28x%29%29+=+sin%28x%29cos%28x%29%284%29\"
\n" ); document.write( "On the left side we need to use the Distributive Property:
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\n" ); document.write( "The reciprocal Trig functions will cancel each other out (since their product is 1):
\n" ); document.write( "\"sin%28x%29+%2B+cos%28x%29sqrt%283%29+=+4sin%28x%29cos%28x%29\"
\n" ); document.write( "If we use the Commutative Property on the second term, out left side is exactly what we wanted it to be:
\n" ); document.write( "\"sin%28x%29+%2B+sqrt%283%29cos%28x%29+=+4sin%28x%29cos%28x%29\"
\n" ); document.write( "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:
\n" ); document.write( "\"sin%28x%29+%2B+sqrt%283%29cos%28x%29+=+2%2A2sin%28x%29cos%28x%29\"
\n" ); document.write( "And we can now substitute in the sin(2x):
\n" ); document.write( "\"sin%28x%29+%2B+sqrt%283%29cos%28x%29+=+2%2Asin%282x%29\"
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