Math help!! it's urgent!! i need to get these answers today!!?

sin(x)+sin(π/2-x)=1

sin(x)+cos(x)=1

(sin(x)+cos(x))^2=1^2

sin^2(x)+2sin(x)cos(x)+cos^2(x)=1

1+2sin(x)cos(x)=1

sin(2x)=0

x=0

cos(2x)=cos(x+x)=cos(x)cos(x)-sin(x)sin(x)

=cos^2(x)-sin^2(x)

sin^2(x)+cos^2(x)=1

cos^2(x)=1-sin^2(x)

substitute cos^2(x)

(1-sin^2(x))-sin^2(x)

1-2sin^2(x)

About the proof of cos2x = 1 - 2* (sinx)^2, I agree with the previous answerer's.

Now, about the equation sinx + sin(-x + 1/2) = 1. As sp33dstix commended, you can use the fact that

sin(a) + sin(b) = 2*sin( (a+b)/2 )*cos( (a-b)/2 )

and write it as 2sin(1/4)*cos(x-1/4) (I think he made a mistake there but the conclusion is the same). The arithmetic value of 2sin(1/4) is about 0.4948 which means that this expression can never reach 1 while x moves in the real numbers.

I noticed in your question that you mention you might "fail the whole semester" so I guess that you might attend a university (or a similar institution) and you might have discussed the so called complex numbers. Inside this set of numbers the above equation can be solved, so I will shortly demonstrate its solution if this is the case (although I would find it a bit strange if your teacher gave you this problem together with the second one - the proof - since the latter is much more elementary than the first one. Maybe he wanted you to prove it using complex arithmetic)

We start from the equation cos(z)=a, assuming that z belongs in the complex numbers, and we ask what is z if a>1. In the complex numbers, there is a very important and basic formula called Euler's formula (see the link for more information). Using this formula, one can get the following relation between the cosine function and the exponential function:

`````````e^(i z) - e^(-i z)```

cos(z)=-------------------- where i is the imaginary unit.

.................... 2........

Using this, cos(z) = a becomes e^(i z) - e^(-i z) = 2 a. Naming e^(i z) as u, this equation can be written as:

u - 1/u = 2 a

So, doing some operations on it, we get:

u^2 - 1 = 2 a u =>

u^2 - 2 a u - 1 = 0

This the usual quadratic equation. It's solutions are(see the second link):

u1 = a + sqrt(a^2 -1) and u2 = a - sqrt(a^2 -1)

(Actually, in the complex domain sqrt is a considered a multi-valued function, so we should not directly talk about the 2 roots. We should instead present the root expression using the complex multivalued function Sqrt and from it deduce that it represents two solutions using a singled valued sqrt function for each, that has only real arguments. But, in order to keep things simple I avoid this here. In each case, both u1 and u2 come out to be the two real positive roots presented above).

Now, remember we use the name u to refer to the quantity e^(i z). So, our initial question about cos(z) = a, has two answers, z1 and z2 that satisfy the two expressions:

e^(i z1) = a + sqrt(a^2 -1)

e^(i z2) = a - sqrt(a^2 -1)

Talking, the natural logarithm in both parts in each expression we conclude that the two solutions are:

z1 = 1 / i * ln ( a + sqrt(a^2 -1) )

z2 = 1 / i * ln ( a - sqrt(a^2 -1) )

Applying this result in the problem of your question, where we have that a = 1/( 2*sin(1/4) ) and z = x - 1/4, we get (after adding 1/4 to both parts):

http://tinyurl.com/complexcos-1

http://tinyurl.com/complexcos-2

(click the links to see the expressions)

The numeric values of these expressions are:

x1=0.25 - 1.32899 i

x2=0.25 + 1.32899 i

I wish you the best of luck with you semester. :-)

For the proof:

cos2x=1-2(sinx)^2

You know that:

sin^2(x) = (1/2)*(1-cos(2x))

Sub it into the right side:

1 - 2*[(1/2)*(1-cos(2x))]

1 - 1 + cos(2x)

Therefore:

LS = RS

Now to solve for x (are you sure it is pi/2 and not 1/2)?:

sin(a) + sin(b) = 2*sin( [a+b]/2 )*cos([a-b]/2)

where:

a = x

b = -x + 1/2

Sub in:

2*sin([x - x - 1/2]/2)*cos([x+x - 1/2]/2)

Becomes:

2*sin(-1/4)*cos(x-1/2)

So this becomes a cos graph with height 2*sin(-1/4).

Now when does this graph equal to one?

I have no idea. Because it never does.

Draw it out to see for yourself. It never equals to 1.

http://www.walterzorn.com/grapher/grapher_e.htm

~Distrubute the numbers: 2a-2b+3a-3b-4a+4b: undergo in strategies it rather is needed distribute the type till now the ( ) to all aspects interior the ( ) and that if the type your are dispensing is a destructive, distribute the destructive sign besides ~Now combine like words: 2a+3a-4a-2b-3b+4b 5a-4a-5b+4b ***a-b*** it rather is your very final answer. listed under are some prepare issues and solutions: ~2(2a-b): Distribute the two so which you have *4a-2b* ~ -6(a-b-2c): Distribute the -6 so as which you have *-6a+6b+12c* i wish that I helped. sturdy success!

cos2x=(cos^2x)-sin^2x

=(1-sin^2x)-sin^2x

=1-2sin^2x