<p>2 2
<li> sin2x=
2</li>
<li><p>1+tan x=</p></li>
<li><p>cos2x=
(1+ln x)
6.e
2 </p></li>
<li><p>y =x-4 domain<strong>range</strong>_
7</p></li>
<li><p>ln e </p></li>
<li><p>ln1</p></li>
<li><p>3(n+10)!/5n!</p></li>
</ol>
<p>Ok sorry see if this is better</p>
<p>never mind i give up. This isn't working sorry about this.</p>
<p>use a graphing calculator- you can answer most of the questions with it</p>
<p>Are you trying to solve these?</p>
<p>Is the first one sinx+cosx=2 ?</p>
<p>(sin x + cos x)^2=2^2
sinx^2 + cosx^2 + 2sinxcosx=4
sin2x=3
x=arcsin3 /2</p>
<p>sin 2x =2
2x = arcsin2
x = arcsin2 /2</p>
<p>I can't really read/understand most of the others.</p>
<p>sin 2x =2
2x = arcsin2
x = arcsin2 /2</p>
<p>x= undefined sin is limited to the range -1 to 1</p>
<ol>
<li>sin squared x not squared(x in not up with 2, if you understand me?)
so sin^2 x + cos^2 x=____</li>
<li>1+ tan^2 x=__</li>
<li>cot^2 x +1=__</li>
<li>sin2x=</li>
<li>cos2x=___
6.this one i never heard of it says eliminate the parameter and write a recatangular equation
x= t^2 + 3
y=2t</li>
<li>ln e^7</li>
<li>e^ln3</li>
<li>find domain and range
y= squareroot(x-4)<br>
(sorry that whole thing is squarerooted,my computer skills aren't too great as you can see)</li>
</ol>
<p>see if that is any better</p>
<ol>
<li>(sin(x))^2 + (cos(x))^2 always = 1</li>
</ol>
<p>2 - 5: These appear to ask for simplified ways to write the LHS</p>
<ol>
<li>You need to write y=f(x) or x=g(y), eliminating t.
Since t^2 = x-3, t = sqrt(x-3)
and y = 2t = 2 sqrt(x-3)
(If you want to be picky, you can argue that this should be
t = +/- sqrt(x-3)
and y = +/- 2 sqrt(x-3)
)</li>
</ol>
<p>Alternatively, y = 2t means t = 0.5 y
and x = t^2 + 3 = 0.25 y^2 + 3</p>
<ol>
<li><p>By definition, ln(e^x) = x
so ln(e^7) = 7</p></li>
<li><p>Set y = e^ln(3) and then take natural log of both sides
ln y = ln(3)
so y = 3 = e^ln(3)</p>
<p>(This is one example of the fact that e^(ln(x)) = x )</p></li>
<li><p>y = sqrt(x-4) or (x-4)^0.5
For y to have meaningful values (i.e. not be an imaginary number), you need x-4 to be >= 0, or x >= 4 . This is the domain.
The range is the set of values for y that you would see for all values of x in the domain. For 4 <= x <= infinity, we have 0 <= y <= infinity; this is the range.</p></li>
</ol>
<p>yeah but how did you know how to simplify for 1-5. I don't even know how you know how to simplify them
thanks for the other answers. I think I get that</p>
<p>For the remaning problems displayed in post#6 above:</p>
<ol>
<li><p>1 + (tan x)^2 = 1 + (sin x / cos x)^2
= (cos x)^2 + (sin x)^2 / (cos x)^2
after mult & dividing 1 by (cos x)^2
= 1 / (cos x)^2 since (sin x)^2 + (cos x)^2 always = 1
= (sec x)^2 by definition</p></li>
<li><p>1 + (cot x)^2 = 1 + (cos x / sin x)^2
= 1 / (sin x)^2 using logic similar to that used in prob.2 above
= (cosec x)^2 by definition</p></li>
<li><p>sin 2x = sin(x + x)
= (sin x)(cos x) + (sin x)(cos x) using formula for sin( a + b)
= 2 (sin x)(cos x) </p></li>
<li><p>cos 2x = cos( x + x)
= (cos x)(cos x) - (sin x)(sin x) using formula for sin( a + b)
= (cos x)^2 - (sin x)^2</p>
<h1>You could rewrite this in several other forms as well.</h1></li>
</ol>
<p>You can Google for 'trigonometric formulas' to find lists of known trig. formulas.</p>
<p>mathwiz, il bandito -
you should avoid shooting yourself in a foot on easy questions.
Equations
sin 2x = 3
and
sin 2x = 2
don't have solutions.</p>
<p>You can also notice that in
sin x + cos x = 2
both sin(x) and cos(x) have to be equal to 1, which is impossible for any x.</p>
<p>==================</p>
<h1>MATH 2c side note.</h1>
<p>sin x + cos x =
sqrt(2) [ (1/sqrt(2)) sin x + (1/sqrt(2)) cos x ] =
sqrt(2) [ sin x cos(pi/4) + cos x sin(pi/4) ] =
sqrt(2) sin(x+pi/4) - pretty useful formula to have in your arsenal.</p>
<p>You can see from this formula that
sin x + cos x <= sqrt(2)
and
[sin x + cos x]{max} = sqrt(2) for x=pi/4 (beside other possible values).</p>
<p>If you want to delve deeper into this trig business, follow
<a href="http://talk.collegeconfidential.com/showthread.php?t=60915%5B/url%5D">http://talk.collegeconfidential.com/showthread.php?t=60915</a></p>
<p>Ya i kinda did mess up there ... oh well</p>
<p>optimizerdad
Ok i understand how you got tan^2 x=(sin x / cos x)^2
but how did you get (cos x)^2 + (sin x)^2 / (cos x)^2
and can you explain the sin(a+b) formula
thanks i guess they expect me to know this but i have never been taught it.
ohh how do you simplify ln1</p>
<p>The answers:</p>
<ol>
<li>1</li>
<li>1+tan^2 x = sec^2 x</li>
<li>cot^2 x + 1 = csc^2 x</li>
<li>sin2x = 2sinxcosx</li>
<li>cos2x = cos^2 x - sin^2 x</li>
<li>Have yet to learn analytic geometry</li>
<li>7</li>
<li>3</li>
<li>Domain: 4 to +infinite, Range: 0 to +infinite</li>
</ol>
<p>I believe the 1st 5 were just basic trig identities.</p>
<p>Florida:
Your school should do a better job of teaching basic stuff like this.. :)</p>
<p>In problem(2), you simply need to give the expression a common denominator, after which you can add all the numerators. To use a numerical example, consider</p>
<p>2 + (1/3)</p>
<p>Multiply the 2 by (3/3); you get</p>
<p>(2)(3/3) + (1/3)
or (6/3) + (1/3)</p>
<p>Since (6/3) and (1/3) have the same denominator, you can add the numerators to get
(6+1)/(3) or 7/3 which is another way of writing 2 + (1/3)</p>
<p>Now go back to problem(2); you have
1 + (sin x / cos x)^2
= 1 + ((sin x)^2) / ((cos x)^2</p>
<p>You need (cos x)^2 as a common denominator, so multiply the '1' by ((cos x)^2) / ((cos x)^2) , just like you multiplied by (3/3) in the numerical example; you then have
(1)((cos x)^2) / ((cos x)^2) + ((sin x)^2) / ((cos x)^2)
or ((cos x)^2) / ((cos x)^2) + ((sin x)^2) / ((cos x)^2)</p>
<p>(cos x)^2 is the common denominator here, so you can add the numerators to get
(((cos x)^2) + ((sin x)^2) ) / ((cos x)^2)</p>
<h2>= 1 / ((cos x)^2)</h2>
<h2>As for the formula for sin(a+b), do a Google search for 'trigonometric formulas', like I'd said earlier; you will see plenty of sites that have summaries of these. I don't think you want to sit and derive each of these here.</h2>
<p>ln(1) = 0, since e^0 = 1 .
In fact log(1) is always 0, whether the base of the log is 10, e or any other positive number. To get an intuitive feel for this, calculate values for
e^1
e^0.1
e^0.01 etc, and notice what happens to the calculated values as the power of e gets closer and closer to 0.</p>
<p>ok i am back
Simplify
1. e^(1+lnx) this is what i got. i don'think it is right
2. ln 1/2
3. (4a^5/3)^3/2 =20^5a^5
4. 4xy^-2/ 12x^(-1/3)y^-5 =x^4y^9/27
5. (5a^2/3)(4a^3/2) =65a^5</p>
<p>what is the range and domain of....
1. y=lnx
2. y=lx+3l - 2
3. y=x^2 if x<0
y=x+2 if 03
4. y=1/x ( i thought this was just y not =0 and x not =0) how do I right that in the right form</p>
<p>ok thanks again</p>
<p>somebody please help me!!!!! I really need explanations school starts in 3 days and this packet is due then. All my friends are stuck also. Shows how well our school teaches. PLEASE</p>
<p>From post#15:</p>
<ol>
<li><p>e^(1 + ln x) = (e^1) * (e^ln x) = (e)(x) {since e^lnx = x; see post#7}</p></li>
<li><p>ln(1/2) = ln(1) - ln(2) = 0 - ln(2) {since ln(1)=0, by definition}</p></li>
<li><p>(4a^5/3)^3/2 {assuming it's not ((4a)^5/3))^3/2}
= ( (2^2) * a^5/3)^ 3/2
= 2^(2 * 3/2) * a^((5/3)(3/2))
= (2^3) * a^5/2
= 8 a^5/2</p></li>
</ol>
<p>More later, if I have the time...</p>