1. Identities, Factorials, Logarithms

2. log 2 (4log 3 9) Prove the equation: IS EQUAL TO ( ) log 2 3 ( 4! ) + 5! 2 ( 4! ) 1 – 2sin 2 θ + sin 4 θ + sin 2 θ cos 2 θ + cos 2 θ

5. 1 – 2sin 2 θ + sin 4 θ + sin 2 θ cos 2 θ + cos 2 θ Now that we have picked the harder side of the equation, we can start simplifying it section by section… Let’s start by working on this section of the equation… ( ) log 2 3 ( 4! ) + 5! 2 ( 4! ) 1 – 2sin 2 θ + sin 4 θ + sin 2 θ cos 2 θ cos θ

6. Rewrite both sine and cosine By only using either sine or cosine, you can go through your work a lot easier Let’s rewrite as much as we could with the problem but first let us try to recognize the identities that were used in the problem: Since there is only sine and cosine through out this identity, we would only be needing the identity : 1 = cos 2 θ + sin 2 θ 1 – 2sin 2 θ + sin 4 θ + sin 2 θ cos 2 θ cos θ

7. First, we should factor out any part of the identity that’s factorable: 1 – 2sin 2 θ + sin 4 θ = ( 1 – sin 2 θ ) 2 (1 – sin 2 θ ) 2 + sin 2 θ cos 2 θ cos 2 θ So now the identity can be expressed as: 1 – 2sin 2 θ + sin 4 θ + sin 2 θ cos 2 θ cos θ

8. Now look at: (1 – sin 2 θ ) 2 Compare it to: 1 = cos 2 θ + sin 2 θ Other than the sin 2 θ and 1, there are no other similarities that can be found in the both identities… “ Teachers love teaching this unit because on the cool manipulations you can do, on the other hand students hate it for that same reason” (1 – sin 2 θ ) 2 + sin 2 θ cos 2 θ cos θ

9. “ Teachers love teaching this unit because on the cool manipulations you can do, on the other hand students hate it for that same reason” What if we try to manipulate one of the identities…. ????? 1 = cos 2 θ + sin 2 θ 1 – sin 2 θ = cos 2 θ + sin 2 θ – sin 2 θ ( cos 2 θ ) 2 (1 – sin 2 θ ) = cos 2 θ (1 – sin 2 θ ) 2 This means that can actually equal to

10. Now the identity has been simplified into: = Notice how the numerator has cosine and the denominator has cosine. This means that we can make these terms reduce… (cos 2 θ ) (cos 2 θ ) + sin 2 θ cos 2 θ cos 2 θ cos 2 θ (cos 2 θ + sin 2 θ ) cos 2 θ (1 – sin 2 θ ) 2 + sin 2 θ cos 2 θ cos θ (cos 2 θ ) 2 + sin 2 θ cos 2 θ cos θ

11. Now all that’s left is cos 2 θ + sin 2 θ which according to the identity 1 = cos 2 θ + sin 2 θ is equal to 1 This simplifies our problem into: + 1 ( ) log 2 3 ( 4! ) + 5! 2 ( 4! )

12. Now it’s time to do the other half it… ( ) log 2 3 ( 4! ) + 5! 2 ( 4! ) Factorials (numbers with an exclamation point) The factorial of a non-negative integer n is the product of all positive integers less than or equal to n. n! = ( n -1 )( n-2 )( n-3 )… until n= 1

13. 3 ( 4! ) + 5! 2 ( 4! ) When this expression is expanded it would be this: 3 * (4 * 3* 2 * 1) + (5 * 4* 3 *2 * 1) 2 * (4 * 3* 2 * 1) So we are left with: (3+ 5)/ 2 = 4 Method #1

14. Method # 2 3 ( 4! ) + 5! 2 ( 4! ) Now let’s try doing it in a shorter way… 3 ( 4! ) + 5 * ( 4! ) 2 ( 4! ) 3 ( 4! ) + 5 * ( 4! ) 2 ( 4! ) 4! ( 3 + 5 ) 2 ( 4! ) = 4

15. log 2 (4) Logarithm (to base b ) of a number x is the exponent y of the power by such that x = by . The value used for the base b must be neither 0 nor 1 (nor a root of 1 in the case of the extension to complex numbers ), and is typically 10, e , or 2. There are three parts of a logarithm… log 3 (81) = 4 base Power Exponent

16. log 2 (4) =x or 2 x = 4 A logarithm is an exponent… A logarithm is an exponent… A logarithm is an exponent… A logarithm is an exponent… A logarithm is an exponent… 2 x =2 2 x = 2 therefore log 2 (4) = 2

17. This half of the equation is done: 1 – 2sin 2 θ + sin 4 θ + sin 2 θ cos 2 θ + cos 2 θ ( ) log 2 3 ( 4! ) + 5! 2 ( 4! ) = 2 = 1 2 + 1 = 3 ( ) log 2 3 ( 4! ) + 5! 2 ( 4! ) 1 – 2sin 2 θ + sin 4 θ + sin 2 θ cos 2 θ cos θ