# Pre-calc Idenities & math Help me understand this please

• Feb 9th 2011, 05:40 AM
mrjameslat
Pre-calc Idenities & math Help me understand this please
Ok I don't understand how to do these 3 types of questions.

1. Verify the identity (Can only change one side, and end result must equal each other)
cos^3Xsin^2X=(sin^2X-sin^4X)cosX

2. Sole the equation.
4cosX=1+2cosX

3. Find the exact values of the sine, cosine, and tangent of the angle by using the sum or difference formula.
345 degrees = 300 degrees + 45 degrees

Thanks for any help.
• Feb 9th 2011, 05:54 AM
skeeter
Quote:

Originally Posted by mrjameslat
Ok I don't understand how to do these 3 types of questions.

1. Verify the identity (Can only change one side, and end result must equal each other)
cos^3Xsin^2X=(sin^2X-sin^4X)cosX

left side ... cosx(cos^2x sin^2x) = cosx[(1 - sin^2x)sin^2x] = cosx(sin^2x - sin^4x)

2. Sole the equation.
4cosX=1+2cosX

subtract 2cosx from both sides ...

2cosx = 1

cosx = 1/2 ... x = pi/3, x = 5pi/3

3. Find the exact values of the sine, cosine, and tangent of the angle by using the sum or difference formula.
345 degrees = 300 degrees + 45 degrees

use the sum/difference identities ...

cos(a+b) = cosa cosb - sina sinb

sin(a+b) = sina cosb + cosa sinb

tan(a+b) = (tana + tanb)/(1 - tana tanb)

in addition, you should be very familiar with the values on the unit circle ...

http://hernandiaz.org/wp-content/upl...unitcircle.gif
• Feb 9th 2011, 07:57 AM
mrjameslat
Thanks for the input.
I guess i can make a paradigm of what you said. Thanks a lot.
:)

however with the last one would this be right

345 degrees = 300 degrees + 45 degrees

cos(300+45) = cos(300) cos(45) - sin(300) sin(45)

sin(300+45) = sin(300) cos(45) + cos(300) sin(45)

tan(300+45) = (tan(300) + tan(45))/(1 - tan(300) tan(45))

and would they equal this.
cos=1/2(√2/2)--√3/2(√2/2)=( √2+√6 )/4

sin=-√3/2(1/2)+1/2(√2/2)=(√2-√6)/4

and how would i do the tangent one? thanks again.
• Feb 9th 2011, 10:51 AM
mrjameslat
Did i do it right?
• Feb 9th 2011, 11:33 AM
skeeter
Quote:

Originally Posted by mrjameslat
however with the last one would this be right

345 degrees = 300 degrees + 45 degrees

cos(300+45) = cos(300) cos(45) - sin(300) sin(45)

sin(300+45) = sin(300) cos(45) + cos(300) sin(45)

tan(300+45) = (tan(300) + tan(45))/(1 - tan(300) tan(45))

and would they equal this.
cos=1/2(√2/2)--√3/2(√2/2)=( √2+√6 )/4 ... ok

sin=-√3/2(1/2)+1/2(√2/2)=(√2-√6)/4 ... ok

and how would i do the tangent one? tanx = sinx/cosx

... you can also check your results w/ a calculator.
• Feb 9th 2011, 11:36 AM
Ackbeet
Quote:

Originally Posted by skeeter
in addition, you should be very familiar with the values on the unit circle ...

http://hernandiaz.org/wp-content/upl...unitcircle.gif

I can get all that information by knowing the 45-45-90 triangle, the 30-60-90 triangle, and the graphs of sin and cosine.
• Feb 9th 2011, 12:13 PM
skeeter
Quote:

Originally Posted by Ackbeet
I can get all that information by knowing the 45-45-90 triangle, the 30-60-90 triangle, and the graphs of sin and cosine.

... and all I need is the 45-45-90 triangle, the 30-60-90 triangle, and the beauty of symmetry. (Wink)
• Feb 9th 2011, 12:17 PM
Ackbeet
Out of curiosity, how would you get the values of sin and cosine on the x and y axes via symmetry?
• Feb 9th 2011, 12:30 PM
skeeter
Quote:

Originally Posted by Ackbeet
Out of curiosity, how would you get the values of sin and cosine on the x and y axes via symmetry?

the picture of the unit circle is color-coded to show that symmetry from quadrant to quadrant.
• Feb 9th 2011, 12:32 PM
Ackbeet
Yeah, but I'm talking about values that are on the x axis, or on the y axis. By symmetry, how do you know that $\displaystyle \cos(\pi/2)=0?$
• Feb 9th 2011, 12:48 PM
skeeter
I guess I would need to remember the basic definition ... $\displaystyle \cos{\theta} = \dfrac{x}{r}$.
• Feb 9th 2011, 12:55 PM
Ackbeet
Ah, so you'd have to remember something in addition to the 45-45-90, the 30-60-90, and symmetry. (Wink)
• Feb 9th 2011, 01:32 PM
mrjameslat

tan(300+45) = (sina/cosa + sinb/cosb)/(1 – sina/cosa*sinb/cosb) =

( -√3/2/1/2 + √2/2/1/2 )/(1- -√3/2/1/2 * √2/2/1/2)=

(-2√3+2√2) / (1+2√3*2√2)

What do I need to do, or what have I done wrong?
thanks
• Feb 9th 2011, 02:04 PM
skeeter
Quote:

Originally Posted by mrjameslat

tan(300+45) = (sina/cosa + sinb/cosb)/(1 – sina/cosa*sinb/cosb) =

( -√3/2/1/2 + √2/2/1/2 )/(1- -√3/2/1/2 * √2/2/1/2)=

(-2√3+2√2) / (1+2√3*2√2)

What do I need to do, or what have I done wrong?
thanks

recheck your values and simplify them before using the sum identity for tangent ...

$\displaystyle \tan(300) = -\sqrt{3}$

$\displaystyle \tan(45) = 1$

$\displaystyle \tan(300+45) = \dfrac{1 - \sqrt{3}}{1 + \sqrt{3}}$

if you decide to rationalize the denominator, the value will simplify to $\displaystyle \sqrt{3} - 2$
• Feb 9th 2011, 03:10 PM
mrjameslat
"work smart, not hard" thanks a lot!