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

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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) About Me  (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 sin%28x%29+=+-3%2F5, we can use the pythagorean trig identity sin%5E2%28x%29%2Bcos%5E2%28x%29+=+1 to determine that cos%28x%29+=+-4%2F5. 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

sin%28A%2BB%29+=+sin%28A%29%2Acos%28B%29%2Bcos%28A%29%2Asin%28B%29

sin%28x%2B5pi%2F6%29+=+sin%28x%29%2Acos%285pi%2F6%29%2Bcos%28x%29%2Asin%285pi%2F6%29

sin%28x%2B5pi%2F6%29+=+%28-3%2F5%29%2Acos%285pi%2F6%29%2B%28-4%2F5%29%2Asin%285pi%2F6%29 Plug in sin(x) = -3/5 and cos(x) = -4/5

sin%28x%2B5pi%2F6%29+=+%28-3%2F5%29%2A%28-sqrt%283%29%2F2%29%2B%28-4%2F5%29%2A%281%2F2%29 Use the unit circle.

sin%28x%2B5pi%2F6%29+=+%283%2Asqrt%283%29%29%2F10-4%2F10

sin%28x%2B5pi%2F6%29+=+%283%2Asqrt%283%29-4%29%2F10

sin%28x%2B5pi%2F6%29+=+%28-4%2B3%2Asqrt%283%29%29%2F10

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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

cos%28A+-+B%29+=+cos%28A%29cos%28B%29+%2B+sin%28A%29sin%28B%29

cos%28x+-+5pi%2F3%29+=+cos%28x%29cos%285pi%2F3%29+%2B+sin%28x%29sin%285pi%2F3%29

cos%28x+-+5pi%2F3%29+=+%28-4%2F5%29%2A%281%2F2%29+%2B+%28-3%2F5%29%2A%28-sqrt%283%29%2F2%29

cos%28x+-+5pi%2F3%29+=+-4%2F10+%2B+%283%2Asqrt%283%29%29%2F10

cos%28x+-+5pi%2F3%29+=+%28-4+%2B+3%2Asqrt%283%29%29%2F10
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 sin%28x%2B5pi%2F6%29+=+cos%28x-5pi%2F3%29 is a trig identity.

The proof is shown in the steps below.

sin%28A%29+=+cos%28pi%2F2-A%29 One of the many trig identities

sin%28x%2B5pi%2F6%29+=+cos%28pi%2F2+-+%28x%2B5pi%2F6%29%29 Plug in A = x+5pi/6

sin%28x%2B5pi%2F6%29+=+cos%28pi%2F2+-+x-5pi%2F6%29

sin%28x%2B5pi%2F6%29+=+cos%283pi%2F6+-+x-5pi%2F6%29

sin%28x%2B5pi%2F6%29+=+cos%28-x-2pi%2F6%29

sin%28x%2B5pi%2F6%29+=+cos%28-x-pi%2F3%29

sin%28x%2B5pi%2F6%29+=+cos%28-%28x%2Bpi%2F3%29%29

sin%28x%2B5pi%2F6%29+=+cos%28x%2Bpi%2F3%29 Use the identity cos(-x) = cos(x)

sin%28x%2B5pi%2F6%29+=+cos%28x%2Bpi%2F3-2pi%29 Use the identity cos(x) = cos(x-2pi)

sin%28x%2B5pi%2F6%29+=+cos%28x%2Bpi%2F3-6pi%2F3%29

sin%28x%2B5pi%2F6%29+=+cos%28x-5pi%2F3%29

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.