onur369's Methods Question Thread :)

[brightsky]

cos(pi - x) = -cos(x) by symmetry. we know that sin^2(x) + cos^2(x) = 1, so cos(x) = sqrt(1 - sin^2(x)) = sqrt(1 - 0.3) = sqrt(0.7).

sin(pi - a) = sin(a) from symmetry. using the above identity, sin(a) = sqrt(1 - cos^2(a)) = sqrt(1 - 0.5) = sqrt(0.5)

Don't you mean
cos(x) = sqrt(1 - sin^2(x)) = sqrt(1 - 0.09) = sqrt(0.91) therefore cos(pi-x)=-sqrt(0.91)?
similarly
sin(a) = sqrt(1 - cos^2(a)) = sqrt(1 - 0.25) = sqrt(0.75)?

oh lol..haha yes. thanks jbebbo!

[Respect]

cos(pi - x) = -cos(x) by symmetry. we know that sin^2(x) + cos^2(x) = 1, so cos(x) = sqrt(1 - sin^2(x)) = sqrt(1 - 0.3) = sqrt(0.7).

sin(pi - a) = sin(a) from symmetry. using the above identity, sin(a) = sqrt(1 - cos^2(a)) = sqrt(1 - 0.5) = sqrt(0.5)

Don't you mean
cos(x) = sqrt(1 - sin^2(x)) = sqrt(1 - 0.09) = sqrt(0.91) therefore cos(pi-x)=-sqrt(0.91)?
similarly
sin(a) = sqrt(1 - cos^2(a)) = sqrt(1 - 0.25) = sqrt(0.75)?

uhh im sorry bro, i still havent understood.. where did u get the formula cos(x) = sqrt(1 - sin^2(x))? and sin(a) = sqrt(1 - cos^2(a)) ... i dont have anything like that in my book, nor do they explain how to do these questions...

[xZero]

trig identity, cos(x)^2 + sin(x)^2 = 1,   cos(x)^2 = 1 - sin(x)^2,   cos(x) = sqrt(1 - sin^2(x))
 cos(x)^2 + sin(x)^2 = 1,   sin(x)^2 = 1 - cos(x)^2,   sin(x) = sqrt(1 - cos^2(x))

[panicatthelunchbar]

hi guys,

If 3+log2(4x)=log2 y, find y in terms of x

[brightsky]

3 + log2(4x) = log2 y
log2(2^3) + log2(4x) = log2 y
log2(8*4x) = log2 y
y = 32x

[panicatthelunchbar]

thanks

[onur369]

Most n00b question to date. sin2x=-1 solve equation between (-n,n) n=pi 
The -1 is confusing me.

[david10d]

The basic angle for that is -pi/2. The rest I'm sure you can work out.

[onur369]

Stuck on another question, that +60 degrees part confuses me, all equations should be from 0-320degrees.

[david10d]

Just think of that as pi/3.

It's just expressed in degrees instead of radians.

[onur369]

Another n00b question. 2sinx+1=b, where b is a positive real number, has one solution in the interval 0,2n?

[luken93]

sin(x) = (b - 1)/2
Thinking about the graph of sin(x), it has a max at (pi/2, 1) and min at (3pi/2, -1)

Thus, (b - 1)/2 =1
b - 1 = 2
b = 3

[Respect]

sin(x) = (b - 1)/2
Thinking about the graph of sin(x), it has a max at (pi/2, 1) and min at (3pi/2, -1)

Thus, (b - 1)/2 =1
b - 1 = 2
b = 3

^
whaa? didnt understand that :/

do u know how to graph y=2sin( l x-pi/2 l ) to one period?

[luken93]

sin(x) = (b - 1)/2
Thinking about the graph of sin(x), it has a max at (pi/2, 1) and min at (3pi/2, -1)

Thus, (b - 1)/2 =1
b - 1 = 2
b = 3

^
whaa? didnt understand that :/

do u know how to graph y=2sin( l x-pi/2 l ) to one period?

I rearranged the equation to make it sin(x) = ....

Now think about the graph of sin(x) from [0, 2pi]

    --
  /     \
/         \          .[2pi, 0]
            \        /
              \     /
                --

For it to have one solution only, it must hit the graph once only. The only points at which this occurs are at the min and max of the graph, ie at [pi/2, 1] and [3pi/2, -1]

Since b is a positive number, then it must be the point pi/2, 1
Solving for this yields b = 3

---------

As for graphing that, draw an imaginary line at x = pi/2
Then graph the function as normal to the right of this line.
Because it is modulus, it will be reflected on this line (think about when x = 0 for example).

[Respect]

its an absolute function, thats not right.