# Trig Identity & Equations simplification

#### nursejason

So I've tried to wrap my head around this concept, and I just read the last two sections in this chapter in order to really get a proper mindset here. Despite reading, there are no examples in the book that pertain to these questions, nor is there anything saying what kind of problem it is and as such I'm not able to look anything up.

Some problems are as follows :
Simplify the given expression :

A) If sin x = 1/3 and 0 < x < pi/2, then sin(pi/4 + x) = ?

The answer to this problem is 4+ sqrt 2 all over 6. And I have no idea how to get there. This section in the chapter mostly discusses sin(A+B) = SinASinB-CosACosB

B) If Sin x = -3/4 and 3pi/2 < x < 2pi then cos (pi/4 + x) = ?

If you don't know how to solve it, but at least know what I could look up to figure out how to get to this solution it would be much appreciated. Final is tomorrow and I'm trying to teach myself most of the concepts as my professor was pretty subpar this semester, and I'm not the best student!

#### chiro

So I've tried to wrap my head around this concept, and I just read the last two sections in this chapter in order to really get a proper mindset here. Despite reading, there are no examples in the book that pertain to these questions, nor is there anything saying what kind of problem it is and as such I'm not able to look anything up.

Some problems are as follows :
Simplify the given expression :

A) If sin x = 1/3 and 0 < x < pi/2, then sin(pi/4 + x) = ?

The answer to this problem is 4+ sqrt 2 all over 6. And I have no idea how to get there. This section in the chapter mostly discusses sin(A+B) = SinASinB-CosACosB

B) If Sin x = -3/4 and 3pi/2 < x < 2pi then cos (pi/4 + x) = ?

If you don't know how to solve it, but at least know what I could look up to figure out how to get to this solution it would be much appreciated. Final is tomorrow and I'm trying to teach myself most of the concepts as my professor was pretty subpar this semester, and I'm not the best student!
Hey nursejason and welcome to the forums.

For both problems you use the addition formulas for angles within trig functions.

So as an example for the first problem you're given sin(x) = 1/3. You are asked to find sin(x + pi/4).

You can simply use the fact that sin(x + y) = sin(x)cos(y) + cos(x)sin(y). By substituting x = the x you have and y = pi/4 we get

sin(x + pi/4) = sin(x)cos(pi/4) + cos(x)sin(pi/4)

Now given that [sin(x)]^2 + [cos(x)]^2 = 1 means cos(x) = SQRT(1 - [sin(x)^2) = SQRT( 1 - 1/9) = SQRT(8/9). But depending on the quadrant cos(x) will either be positive or negative, which we will come back to.

Ok so cos(x) = (+ or -) SQRT(8/9) = (+,-) 2/3 x SQRT(2). Plug this into your sin(x + pi/4) noting that sin(pi/4) = cos(pi/4) = 1/SQRT(2) we get

sin(x + pi/4) = 1/3 x 1/SQRT(2) + 2/3 x SQRT(2) x 1/SQRT(2)
= 1/3 x 1/SQRT(2) + 2/3
= (3x1 + 3xSQRT(2)x2)/9xSQRT(2)
= (3 + 6SQRT(2)) / (9xSQRT(2))
= (1 + 2SQRT(2)) / 3SQRT(2)

Multiply this by SQRT(2)/SQRT(2) and you get

= SQRT(2)/SQRT(2) x (1 + 2SQRT(2))/(3xSQRT(2))
= (SQRT(2) + 4) / 6

Which is what the book gives.

If you're wondering about why I used + instead of -, look at the unit circle for the range which the angle is in. For example in 1st quadrant both are positive, but in 2nd cos is negative, 3rd sin negative and so on.

#### nursejason

I may or may not be able to kiss you. I am a little lost at one part though, yet I understand most of the rest.

Ok so cos(x) = (+ or -) SQRT(8/9) = (+,-) 2/3 x SQRT(2). Plug this into your sin(x + pi/4) noting that sin(pi/4) = cos(pi/4) = 1/SQRT(2) we get
I lost you a little bit here. I am plugging in the +- 2/3 rt2 where exactly? I wrote this down and tried to figure it out a few ways but got a little lost.

Similarly I don't know how you got from the top line, to the one following. It's a lot more confusing than I would have imagined trying to convert keyboard text to what you're used to seeing in a text or on paper :P

= 1/3 x 1/SQRT(2) + 2/3
= (3x1 + 3xSQRT(2)x2)/9xSQRT(2)
= (3 + 6SQRT(2)) / (9xSQRT(2))
= (1 + 2SQRT(2)) / 3SQRT(2)
Like I said, I understand the top line just fine. But I'm not sure how the rest flows.

Thanks so much for your help thus far!

#### chiro

I may or may not be able to kiss you. I am a little lost at one part though, yet I understand most of the rest.

I lost you a little bit here. I am plugging in the +- 2/3 rt2 where exactly? I wrote this down and tried to figure it out a few ways but got a little lost.
You're substituting + 2/3 ROOT(2) for cos(x). That was figured out by using the fact that
sin^2(x) + cos^2(x) = 1. The + is because of the quadrant that x is in.

Similarly I don't know how you got from the top line, to the one following. It's a lot more confusing than I would have imagined trying to convert keyboard text to what you're used to seeing in a text or on paper :P
So I said that sin(x + pi/4) = sin(x)cos(pi/4) + cos(x)sin(pi/4)

sin(pi/4) = cos(pi/4) = 1/ROOT(2). sin(x) = 1/3, cos(x) = + 2/3 x ROOT(2)

Like I said, I understand the top line just fine. But I'm not sure how the rest flows.

Thanks so much for your help thus far!
Going from what I said above the rest is just collecting terms.

When I wrote a SQRT(2)/SQRT(2) all I did was multiply everything by 1 (which doesn't change anything). That basically gets rid of the square root on the denominator and when you collect all the terms together you get the answer.

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