SOLUTION: Hi ! I have a couple of questions that I am stuck on! First: Express sin(x^2+1) in terms of sin(x^2) and cos(x^2) second: Express tan(x+2) in terms of tan(x). On

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: Hi ! I have a couple of questions that I am stuck on! First: Express sin(x^2+1) in terms of sin(x^2) and cos(x^2) second: Express tan(x+2) in terms of tan(x). On      Log On


   

Question 163546: Hi !
I have a couple of questions that I am stuck on!
First:
Express sin(x^2+1) in terms of sin(x^2) and cos(x^2)
second:
Express tan(x+2) in terms of tan(x).
One other question I have, if you are given coordinates is it possible to work out the equation of a straight line from the coordinates?
e.g. my question is:
Find the equation of a straight line which passes through the points (-1, -5) and 3,3)
Really appreciate any help i can get! Thanks!!
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
1.sin%28a%2Bb%29=sin%28a%29cos%28b%29%2Bcos%28a%29sin%28b%29
sin%28x%5E2%2B1%29=sin%28x%5E2%29cos%281%29%2Bcos%28x%5E2%29sin%281%29
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.
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2.tan%28a%2Bb%29=%28tan%28a%29%2Btan%28b%29%29%2F%281-tan%28a%29tan%28b%29%29
tan%28x%2B2%29=%28tan%28x%29%2Btan%282%29%29%2F%281-tan%28x%29tan%282%29%29
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3.Yes, you can calculate the slope from the two points.
Then use the point-slope form of a line.
First, the slope,
m=%28y%5B2%5D-y%5B1%5D%29%2F%28x%5B2%5D-x%5B1%5D%29%29
m=%283-%28-5%29%29%2F%283-%28-1%29%29%29
m=8%2F4%29
m=2
Now that you have the slope, use the point-slope form,
y-y%5Bp%5D=m%28x-x%5Bp%5D%29
You can use either point.
y-3=2%28x-3%29
y-3=2x-6
y=2x-3
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.
.
Here's graphical verification.