# Thread: Need sum and difference on trigonomtery

1. ## Need sum and difference on trigonomtery

Hi,

I'm getting close to the right answer, but one little thing is off. I think I'm missing a negative somewhere...

"Find the exact value for each of the following under the given conditions:
(a) sin(alpha + beta) ... etc...

Sin alpha = 5/13 , -3pi/2 < alpha < -pi ; Tan beta = -sqrt of 3 , pi/2 < beta < pi
Here's my work:

I make a couple of pictures to find the missing side. For alpha I get a triangle in quadrant II. I get the missing, base, side is equal to 12 after using the Pythagorean theorem. The 12 is also negative because of the quadrant.

For beta I get another triangle in quadrant II. I get the missing, hypotenuse, side is equal to 2.

I then plug the numbers into the sum and difference formula for Sin(alpha + beta) which is sin alpha cos beta + cos alpha sin beta

I get (5/13) (-1/2) + (-12/13) (sqrt3/2)

= (-5/26) + (-12sqrt3/26)

= -5 - 12sqrt3 all over 26.

The answer should be 5 + 12sqrt3 all over 26.

Where are my negatives coming from? Did I use the wrong quadrant or did I miss something in the work?

Thanks,

-Tom

2. ## Double angle formula help as well...

This one is giving me issues too...

Use the information given about the angle theta, 0 < theta < 2pi, to find the exact value of
(a) sin(2theta) ... etc...

cos theta = - sqrt6/3, pi/2 < theta < pi
I find the missing side... -sqrt6 + y^2 = -3^2

-6 + y^2 = -9
+6 +6

y^2 = -3

y = -sqrt3

Then I plug them into the double angle formula... 2 sin theta cos theta.

2 (sqrt3/3)(- sqrt6/3)

Then I get 2( - sqrt18/3)

The answer should be -2 sqrt2/3 ... Did I just make a simple mistake or something?

Thanks again,

-Tom

3. Originally Posted by TomCat
Hi,

I'm getting close to the right answer, but one little thing is off. I think I'm missing a negative somewhere...

Here's my work:

I make a couple of pictures to find the missing side. For alpha I get a triangle in quadrant II. I get the missing, base, side is equal to 12 after using the Pythagorean theorem. The 12 is also negative because of the quadrant.

For beta I get another triangle in quadrant II. I get the missing, hypotenuse, side is equal to 2.

I then plug the numbers into the sum and difference formula for Sin(alpha + beta) which is sin alpha cos beta + cos alpha sin beta

I get (5/13) (-1/2) + (-12/13) (sqrt3/2)

= (-5/26) + (-12sqrt3/26)

= -5 - 12sqrt3 all over 26.

The answer should be 5 + 12sqrt3 all over 26.

Where are my negatives coming from? Did I use the wrong quadrant or did I miss something in the work?

Thanks,

-Tom
Your work looks good to me.

-Dan

4. Originally Posted by TomCat
2 (sqrt3/3)(- sqrt6/3)

Then I get 2( - sqrt18/3)
Careful.
2 (sqrt3/3)(- sqrt6/3) = -2*sqrt(18)/9

Now, sqrt(18) = sqrt(2*9) = 3*sqrt(2)

So
2 (sqrt3/3)(- sqrt6/3) = -2*sqrt(18)/9 = -2*3*sqrt(2)/9 = -2*sqrt(2)/3

-Dan

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### sin alpha is 1÷2 cos beta is √3÷2 find alpha beta

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