<PRE><B>Question 236921</B>
First, we&#39;ll need the following formulas for these two problems:
{{{cos(2x) = cos^2(x) - sin^2(x)}}}
{{{sin(2x) = 2sin(x)cos(x)}}}&lt;br&gt;
To become skilled at using these (and the other Trig. formulas):&lt;ul&gt;&lt;li&gt;Look at the x&#39;s as only placeholders. The x&#39;s can be anything. So, using the first of the two properties above:
{{{cos(2y) = cos^2(y) - sin^2(y)}}}
{{{cos(20) = cos^2(10) - sin^2(10)}}}
{{{cos(2x+2y) = cos^2(x+y) - sin^2(x+y)}}}
{{{cos((1/5)x) = cos^2((1/10)x) - sin^2((1/10)x)}}} (Since 1/5 = 2*(1/10))&lt;/li&gt;&lt;li&gt;The coefficients in the formula can often be &quot;matched&quot; by use of factoring. (Both of your problems will need this so after seeing their solutions I hope you understand this tip better.)&lt;/li&gt;&lt;/ul&gt;
Now let&#39;s try your problems.
{{{7cos^2(theta/9)-7sin^2(theta/9)}}}
We can use the cos(2x) formula on this because:&lt;ul&gt;&lt;li&gt;The formula has cos^2 - sin^2 and so does your expression.&lt;/li&gt;&lt;li&gt;The formula has the same arguments (x) for cos^2 and sin^2 and so does your expression (theta/9).&lt;/li&gt;&lt;li&gt;The formula has the same coefficients (1) for cos^2 and sin^2 and so does your expression (7).&lt;/li&gt;&lt;/ul&gt;
This is how you figure out if a formula can be used. Before we actually go ahead and use the formula we need to &quot;match&quot; the coefficients. This is done by factoring (sometimes creatively).&lt;br&gt;
Your coefficients are 7&#39;s and we want 1&#39;s. So factor out a 7:
{{{7(cos^2(theta/9) - sin^2(theta/9))}}}
Inside the parentheses we have the exact pattern of the cos(2x) forumula. So we can replace that expression with cos(2*theta/9):
{{{7(cos(2*theta/9))}}}
And we can remove the outer parentheses:
{{{7cos(2*theta/9)}}}&lt;br&gt;
{{{sin(theta/6)cos(theta/6)}}}
We can use the sin(2x) formula on this because:&lt;ul&gt;&lt;li&gt;The formula has sin * cos and so does your expression.&lt;/li&gt;&lt;li&gt;The formula has the same arguments (x) for cos and sin and so does your expression (theta/6).&lt;/li&gt;&lt;li&gt;The formula has a coefficient of 2 and your expression can have a coefficient of 2 with some creativity.&lt;/li&gt;&lt;/ul&gt;
Here&#39;s how to get the coefficient we need. As you know, multiplying something by one does not change it. To get the 2 we need as a coefficient, we need to create a 1, as a product, which has a two in it. Since the product of reciprocals is always 1, we need (1/2)*2. So we multiply by this giving:
{{{(1/2)*2*sin(theta/6)cos(theta/6)}}}
Using the Associative Property we can group and separate the expression that matches the sin(2x) pattern:
{{{(1/2)*(2*sin(theta/6)cos(theta/6))}}}
Now we can replace the pattern with sin(2*theta/6):
{{{(1/2)*sin(theta/3)}}}</PRE>