Make up your OWN identity. Start with a trigonometric expression, and apply substitutions and algebraic processes to create equivalent expressions.

Justify each step. Verify your identity graphically and algebraically.

I have already made up my own identity witch is

1/4 sin^2 (2x) = sin^2(x) * (1-sin^2 (x))

here is what I am expecting you to answer me the following identity is the teachers identity please dont copy just see so you can have an idea on how to work on it :

Making an Identity 5;va + {Cm U): {Ol 3C, : SITWJf &,C'l :3 : C Elf-i .JlJ C033: C094. {3 , w} tonl++ajz 3' \\4- W) C031 C033] \\ W C031 083 7' 3t\" 21+ (3033(- C038) The rst thing I did when creating my identity was choosing which trigonometric ratio to explore. I decided on tangent because it goes back to the relationship between sine and cosine. In my case, | explored tan x and tan y. This is because this activity developed an understanding of identities between addition and subtraction. The best way for me to approach those types of identities was to have two variables; x and y. In the photo above, I described what tan x and tan y would equal. Then, I substituted their simplied values to replace \"tan". Hence why... tan x + tan y = {sin xicos x) + (sin yicos y) In this case, I had to add two fractions. Except, they did not have a common denominator, so I algebraically multiplied each term by the opposite denominator so that I can add the fractions. I multiplied the rst term (sin xfcos x] by cos y. Then, I multiplied the second term, (sin yioos y), by cos x As a result I had w {.05 1 (0.5 y In this activity, I learned the addition trig identity which is. .. sinAcosB + sinBcosA = sin(A + B) This trig identity Is applicable to the numerator of w where 1-: represents A' and y EDS I ['05 _'|-' represents\" 3'. Hence why I substituted 'sinlx + y)\" as sin 2: cos y + sin y cos 3:. sin {x+y'] EOSICOS }' So, tan x + tan y = {dost {cowl-5:19;\" \"M c0571 C63 5 1 rs i COSJC 035 .. : 'h _ Co$1COS\\5 After determining the rst part of my new identity, I continued to build up on the connection between tangent ratios and sine and cosine. So, I explored (tan 1: - tan y). This is similar to what I did in the rst step, except instead of addition, the ratios are subtracted. In this case, I had to subtract two fractions. Except, they did not have a common denominator, so I algebraically multiplied each term by the opposite denominator so that I can subtract the fractions. | multiplied the rst term, (sin xfcos x], by cos y. Then, I multiplied the second term, sin x to: JP sin: y cos 1' {sin yrcos y}, by cos x. As a result, i had mum\" In this activity, i learned the subtraction trig identity which is... sinAcosB - sinBcosA = sin[A - B) This trig identity Is applicable to the numerator of w where x represents 'A' and y ['05 I L'l'i'Sy represents\" 3'. Hence why I substituted 'sin(x - y)\" as stnx cos y sot 3: cos x. sin. (Jry} So, tan x tan y = mum\" The third step I did was to create the rst side of my identity. Let's say that this will be the RS. To put it simply, I used the two Expressions that l was exploring and decided to divide them by each other. After that, l substituted the simplied versions of them that I had calculated earlier. Recall that when an expression is divided by another expression, it is essentially multiplied by its reciprocal. So. I multiplied Mby the reciprocal of mm\" which is comm cos x cos 1; cos x cos y sin [xrj ' During the multiplication process, I noticed that the like terms (cos x cos y) cross each other out. So, all l was left with was M sin. Ex+y] ' Since cannot be simplied further, then I have concluded that my identity is Although this is proven algebraically, it can also be proven by Desmos' graph calculation system... + I" 1}