SOLUTION: Suppose sin(x) = -3/5 and tan(x) > 0. Find sin(x + 5𝜋/6) and cos(x

SOLUTION: Suppose sin(x) = -3/5 and tan(x) > 0. Find sin(x + 5𝜋/6) and cos(x - 5𝜋/3).

Question 1198267: Suppose sin(x) = -3/5 and tan(x) > 0.
Find sin(x + 5𝜋/6) and cos(x - 5𝜋/3).
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

Part 1) Find sin(x+5pi/6)

sin(x) < 0 and tan(x) > 0
This places angle x in quadrant III, aka the southwest quadrant.

Because , we can use the pythagorean trig identity to determine that . Recall that cosine is negative in Q3.

Use this trig identity
sin(A+B) = sin(A)*cos(B)+cos(A)*sin(B)
in this context
A = x
B = 5pi/6

Plug in sin(x) = -3/5 and cos(x) = -4/5

Use the unit circle.

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Part 2) Find cos(x - 5pi/3)

This time we'll be using this trig identity
cos(A - B) = cos(A)cos(B) + sin(A)sin(B)
we have
A = x
B = 5pi/3

We get the same answer as before.

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Bonus Optional Section:

It's not a coincidence that we got the same result each time.
It turns out that is a trig identity.

The proof is shown in the steps below.

One of the many trig identities

Plug in A = x+5pi/6

Use the identity cos(-x) = cos(x)

Use the identity cos(x) = cos(x-2pi)

This confirms that sin(x+5pi/6) and cos(x-5pi/3) are the same thing, but in different forms of course.
It's like saying how x+x is the same as 2x.

Cosine is a phase-shifted version of sine (hence the name "cosine" means "cofunction of sine").
If you were to start with sin(x) and apply a phase shift of 5pi/6 units to the left, then you would end up with sin(x+5pi/6)

Now if you were to start with cos(x), and apply a phase shift of 5pi/3 units to the right, then you'd get to cos(x-5pi/3)

Both of these result functions land on the same exact curve.
I recommend using either Desmos or GeoGebra to interact with these curves as described above.

A non-visual approach to "seeing" how the curves are the same is to generate a table of values. You should find that both sin(x+5pi/6) and cos(x-5pi/3) produce the same output for any given x input.
This of course does not constitute a proof (use the steps shown at the top of this section for the actual proof), but rather is a numerical example to help cement the idea of what's going on.

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