Unsure if this is a trig identity or calculus

[Arnoldjavs3]

Homework Statement

$$sinx - cosx = 1/3$$
solve for $$sin(2x)$$

Homework Equations

$$sin^2x + cos^2x = 1$$
$$sin2x = 2cosxsinx$$

The Attempt at a Solution

I think you can square both sides and get:
$$sin^2x - cos^2x = 1/9$$
But how can I use this information to solve for sin2x? Is there a relation/identity between sin2x and sin^2x?

But I am unsure of what to do proceeding this point, is there any calculus needed here? It's in my Calculus review problems so I am confused.

 

[Mark44]

Homework Statement

$$sinx - cosx = 1/3$$
solve for $$sin(2x)$$

Homework Equations

$$sin^2x + cos^2x = 1$$
$$sin2x = 2cosxsinx$$

The Attempt at a Solution

I think you can square both sides and get:
$$sin^2x - cos^2x = 1/9$$

How did you get this? I hope you didn't do just this: $(sin(x))^2 - (cos(x))^2 = (1/3)^2$. If so, that's not at all valid.

In general $(a + b)^2 \neq a^2 + b^2$.

But how can I use this information to solve for sin2x? Is there a relation/identity between sin2x and sin^2x?

But I am unsure of what to do proceeding this point, is there any calculus needed here? It's in my Calculus review problems so I am confused.

 

[vela]

Hint: $(a+b)^2 \ne a^2 + b^2$.

 

[ehild]

$$sinx - cosx = 1/3$$
I think you can square both sides and get:
$$sin^2x - cos^2x = 1/9$$

That is utterly wrong ! How do you square the difference of two terms? What is (a-b)(a-b)?

 

[Arnoldjavs3]

Sigh it's just annoying having to deal with something i did several years ago in a calculus course. I guess I should've remembered the basic factoring property.

Anyway, $$sin^2x + cos^2x - 2sinxcosx = 1/9$$
and I think $$sin^2x + cos^2x = 1$$ so I believe it'll end up being after simplification
$$1-sin2x = 1/9$$
and the answer I got is $$8/9$$. The answer is unfortunately not given for this question.

 

[vela]

Well, you can take solace in the fact that you got the right answer. I've attached a plot of $y=\sin x - \cos x$, $y=1/3$, $y=\sin 2x$, and $y=8/9.$ You can see whenever $\sin x - \cos x$ intersects the line $y=1/3$, the graph of $y=\sin 2x$ intersects the line $y=8/9$.

 

[Arnoldjavs3]

Just out of curiosity is trig identities something I should feel comfortable with? As far as math courses that I'm taking in my programs i believe I take up to calc3, linear algebra and discrete math

 

[Mark44]

Just out of curiosity is trig identities something I should feel comfortable with?

Absolutely. And not just "comfortable with." You should have some of them memorized, including the so-called Pythagorean identities (sin^x(x) + cos²(x) = 1, and the other two involving tan(x), sec(x) and cot(x), csc(x)), the sum identities, the double angle identities, and possibly the half-angle identities.

As far as math courses that I'm taking in my programs i believe I take up to calc3, linear algebra and discrete math

The trig identities won't be used much in discrete math, but they are used to some extent in linear algebra and quite a lot in calculus.

Also, the mistake you made in the first post is less about trig, and more about basic algebra properties: when you apply some operation to one side of an equation, you have to apply the same operation to the other side. In this case, squaring one side of an equation is not the same as squaring the individual terms. What you did seems to be a misapplication of the distributive law,