• Mar 15th 2010, 06:22 PM
purplec16
$tan 60 + tan 225$

$\frac{tan60+tan225}{1-tan60 tan225}$

$\frac{\sqrt{3}+1}{1-\sqrt{3}\times1}$

$\frac{(\sqrt{3}+1)(1+\sqrt{3})}{1-\sqrt{3}}$

$\frac{2\sqrt{3}+4}{-2}$

The answer is suppose to be $-2-\sqrt{3}$

and im not gettin it can anyone help me?

Another question, What is the difference between $tan60+tan 225$ and $tan(60+225)$
• Mar 15th 2010, 06:47 PM
sa-ri-ga-ma
(2*sqrt3 + 4)/-2 = 2*sqrt3/-2 + 4/-2 =.........?
And
tanA + tanB is not equal to tan(A+B)
tan(A+B) = (tanA+tanB)/[1-tanA*tanB]
• Mar 15th 2010, 06:51 PM
TKHunny
1) "The answer is supposed to be..."

That is what you have. Try some algebra. There is a reason why you studied "simplification" over and over and over.

2) What is the difference.

Simply, you do not understand the concept fo funciton notation.

f(a) is a function operating on a value 'a'.
f(b) is a function operating on a value 'b'.
f(a+b) is a function operating on the value 'a + b'.
f(a) + f(b) is ths sum of a function operating on a value 'a' and that same function operating on a value 'b'.

I am not encouraged that 1) You started the problem with the wrong expression and 2) you only got so far as to wonder "they give different answers". You really should start understanding what you are doing. One day you will discover the mathematics is not just a class to get through.
• Mar 15th 2010, 06:52 PM
purplec16
I know how to do the question, Thanks alot I understand now I was just kind of confused
• Mar 15th 2010, 07:27 PM
purplec16
I have another question that I really need help with

If $\alpha$and $\beta$ are acute angles such that $cos\alpha=\frac{4}{5}$ and $tan\beta=\frac{8}{15}$
Find:
$sin(\alpha+\beta)$ and $cos(\alpha+\beta)$

Which quadrant contains $\alpha+\beta$?
• Mar 15th 2010, 07:29 PM
hi purplec16,

x(60)+x(225)=x(60+225)

That's algebraic multiplication.

Tan(angle) is a "trigonometric function".
So you must study the difference.
• Mar 15th 2010, 07:43 PM
Quote:

Originally Posted by purplec16
I have another question that I really need help with

If $\alpha$and $\beta$ are acute angles such that $csc\alpha=\frac{4}{5}$ and $tan\beta=\frac{8}{15}$
Find:
$sin(\alpha+\beta)$ and $cos(\alpha+\beta)$

Which quadrant contains $\alpha+\beta$?

You will need to work these out using Pythagoras' Theorem
and also the following trigonometric identities

$cos(A+B)=cosAcosB-sinAsinB$

$sin(A+B)=sinAcosB+cosAsinB$

To start, draw 2 right-angled triangles.

For the first $csc\alpha=\frac{4}{5}=\frac{1}{sin\alpha}$

Therefore $sin\alpha=\frac{5}{4}=\frac{opposite}{hypotenuse}$

So there is something wrong here as the opposite cannot be longer than the hypotenuse
since the hypotenuse is the longest side of a right-angled triangle.

Then, when you use Pythagoras' theorem, you find the 3rd side
and you can write $cos\alpha$

Do the same with the 2nd triangle, using $tan\beta=\frac{8}{15}=\frac{opposite}{adjacent}$

Again you get the 3rd side of that to help you write $cos\beta$ and $sin\beta$
• Mar 15th 2010, 07:58 PM
purplec16
sorry it is [tex]cos\alpha[\math], I see why cos has to do with it, but I dont see how tan comes into play?
• Mar 15th 2010, 08:03 PM
ok,

just use $cos\alpha=\frac{4}{5}=\frac{adjacent}{hypotenuse}$

Draw the triangle.

Pythagoras' theorem gives $(opposite)^2+(adjacent)^2=(hypotenuse)^2$

Use this to find the length of the 3rd side.

Then write $sin\alpha$

Finally use the trigonometric identity.

You need $sin\beta$ and $cos\beta$ also.

You get these when you have the 3rd side of the 2nd triangle.

$tan\beta$ gives you 2 of the 3 sides of that one.

Pythagoras theorem gets you the 3rd.

Then you can write $cos\beta$ and $sin\beta$

before using the trigonometric identity.
• Mar 16th 2010, 12:14 AM
Hello purplec16
Quote:

Originally Posted by purplec16
I have another question that I really need help with

If $\alpha$and $\beta$ are acute angles such that $cos\alpha=\frac{4}{5}$ and $tan\beta=\frac{8}{15}$
Find:
$sin(\alpha+\beta)$ and $cos(\alpha+\beta)$

Which quadrant contains $\alpha+\beta$?

For future reference, it's more helpful if you post additional questions as a new thread. (See Forum Rules, #14)

• Mar 16th 2010, 05:29 PM
purplec16
Quote:

ok,

just use $cos\alpha=\frac{4}{5}=\frac{adjacent}{hypotenuse}$

Draw the triangle.

Pythagoras' theorem gives $(opposite)^2+(adjacent)^2=(hypotenuse)^2$

Use this to find the length of the 3rd side.

Then write $sin\alpha$

Finally use the trigonometric identity.

You need $sin\beta$ and $cos\beta$ also.

You get these when you have the 3rd side of the 2nd triangle.

$tan\beta$ gives you 2 of the 3 sides of that one.

Pythagoras theorem gets you the 3rd.

Then you can write $cos\beta$ and $sin\beta$

before using the trigonometric identity.

Why does quandrant I contain $\alpha+\beta$
• Mar 16th 2010, 06:29 PM
It isn't a matter of figuring out which quadrant the angle $(\alpha+\beta)$ is in, purplec.

The question does not ask you to find the angle.

It's asking for $sin(\alpha+\beta)$ and $cos(\alpha+\beta)$

Are you able to use Pythagoras theorem to write

$Sin\alpha$

$cos\alpha$

$sin\beta$

$cos\beta$ ?
• Mar 16th 2010, 06:31 PM
purplec16
Yes I am i got the question but it ask the quadrant of alpha + beta?
• Mar 16th 2010, 06:39 PM
Ok,

well, once you have found out what $cos(\alpha+\beta)$
and $sin(\alpha+\beta)$ are

you find out the quadrant of the angle using the following...

$sin(angle)$ gives the vertical co-ordinate of a point on the unit circle circumference, centred at (0,0).

$cos(angle)$ gives the horizontal co-ordinate.

Hence,

sin + and cos + is the 1st quadrant 0-90 degrees

sin + and cos - is the 2nd quadrant 90-180 degrees

sin - and cos - is the 3rd quadrant 180-270 degrees

sin - and cos + is the 4th quadrant 270-360 degrees