Melody

The calc on the home page does do that.   

Just press the equal sign a second time and it will display in scientific notation.    

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The calc on the forum won't do this though  

What is I+1=x I-1=x I=??????

huh??  

What is the question ???

Hi Vest4R

The question is

sin(theta + pi/4) = 1/2

now (theta + pi/4) is an angle so I could substitute    alpha

\(sin\;\alpha=\frac{1}{2}\)

A lot of people just memorize that sin30 degrees=1/2 but I memorize the triangle that it come from. 

I draw an equilateral triangle with sides 2 units and cut it in half.

The right angle triangle thus formed will be this:

Note that 30 degrees is   pi/6 radians

so

\( sin \frac{\pi}{6}=\frac{1}{2}\\ \alpha=\frac{\pi}{6}\\ so\\ \theta+\frac{\pi}{4}=\frac{\pi}{6}\\\)

Does that fully answer your question?    

If you ask more questions you may be advised to private message me this address cause otherwise I may not see it. :)

Start by adding 2 to both sides.

Then divide both sides by -5

sin(theta + pi/4) = 1/2

\(sin(\theta + pi/4) = 1/2\)

Looking at just the acute angle.

\(sin\frac{\pi}{6}=\frac{1}{2}\)

sine is positive in the first and second quads so ...

\(\theta+\frac{\pi}{4}=\frac{\pi}{6},\quad \frac{5\pi}{6},\quad \frac{13\pi}{6}\\ \theta+\frac{3\pi}{12}=\frac{2\pi}{12},\quad \frac{10\pi}{12},\quad \frac{26\pi}{12}\\ \theta=\frac{2\pi}{12}-\frac{3\pi}{12},\quad \frac{10\pi}{12}-\frac{3\pi}{12},\quad \frac{26\pi}{12}-\frac{3\pi}{12}\\ \theta=\qquad \frac{-1\pi}{12},\qquad \frac{7\pi}{12},\qquad \frac{23\pi}{12}\\ but\;\;0\le\theta\le2\pi\\ so\\ \theta= \frac{7\pi}{12},\qquad \frac{23\pi}{12}\\ \)

That is an excellent explanation HSC   

2 sets of ordered pairs for 7x-8y=0

x=0, y= 0  (0,0)

x=8, y=7   (8,7)

It is undefined.

You cannot divide any number by 0

how about 

7*8 = 7*4*2 = 28 doubled = 56

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I was very stubborn when I was a kid and I never learned my tables properly.

To muliply by 8 I always doubled then doubled again (I still use this method)

7*8 = 7, 14, 56 bingo!    LOL

sin^4 2 theta - 2 sin^2 2 theta = -1

\(sin^4 2 \theta - 2 sin^2 2 \theta = -1\\ (sin^2 2\theta)^2 - 2 (sin^2 2\theta )= -1\\ (sin^2 2\theta)^2 - 2 (sin^2 2\theta )+1=0\\ \text{let }x=sin^22\theta\\ x^2 - 2x+1=0\\ (x-1)(x-1)=0\\ x=+1\\ so\\ sin^22\theta=1\\ sin2\theta=\pm1\\ 2\theta=\frac{\pi}{2},\;\;\frac{3\pi}{2},.......\\ 2\theta=\frac{\pi}{2}+n\pi \qquad n\in Z\;\;(n \;is\;an\; integer)\\ \theta=\frac{\pi}{4}+\frac{n\pi}{2} \qquad n\in Z\\ \theta=\frac{\pi(1+2n)}{4} \qquad n\in Z\\ \)

check:

\(sin^4 2 \theta - 2 sin^2 2 \theta = -1\\ LHS\\ =\left(sin \frac{\pi(1+2n)}{2}\right)^4 - 2 \left(sin \frac{\pi(1+2n)}{2}\right)^2\\ =\left(\pm1\right)^4 - 2 \left(\pm1\right)^2\\ =1 - 2 *1\\ =-1\\ =RHS\)

check: