Question

On the Problem Set 1: Question 1A - 6 b)
How did they get there ?
Here's the link :
- Problem: http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-a-definition-and-basic-rules/problem-set-1/MIT18_01SC_pset1prb.pdf
- Solution: http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-a-definition-and-basic-rules/problem-set-1/MIT18_01SC_pset1sol.pdf

Answer 1

If you set x = 0, then you have sin(x) - cos(x) = (0) - (1) = -1.
And you have a amplitude sqrt(2), so at x = 0 your y = \[-\frac{1}{\sqrt2} = -\frac{\sqrt2}{2}\].
Looking at the unit circle you can see that sin(-pi/4) give you this value.

At least that is how I did it.

Answer 2

Use the identity sin (A-B) = sin A cos B - cos A sin B
We want\[\sin x-\cos x=(\sin x) \times1-(\cos x)\times1\] There is no place where sin and cos both equal 1, sin and cos are equal at pi/4

Answer 3

\[\sqrt{2}\times \sin (\pi/4) = 1\]\[\sqrt{2}\times \cos (\pi/4) = 1\]

Answer 4

So we multiply sin(A-B) by √2 and A=x and B=π/4

Answer 5

Thank a lot for your help!