Question

Can someone please explain to me how they get from steps 3 to 4 on this identity?!?

http://aplus.escholaracademy.net/R85Statics/media/pictures/trigonometry/sum%20difference%20identities/study/page15/981729422359/1366x768/981729422359.png

Answer 1

here is the problem with the triangles

Answer 2

*cough* what are you finding anyway?

Answer 3

hmm ohhh do hh the cosine of 105 degrees... hmm

Answer 4

Ya im going through sum and difference identities right now but i just dont know how they got the last step from the one before it

Answer 5

right

Answer 6

I dont know how they simplified the 3rd step

Answer 7

\(\bf \left(\cfrac{\sqrt{2}}{2}\cdot \cfrac{1}{2}\right)-\left(\cfrac{\sqrt{2}}{2}\cdot\cfrac{\sqrt{3}}{2}\right)\implies
\left(\cfrac{\sqrt{2}}{4}\right)-\left(\cfrac{\sqrt{2\cdot 3}}{4}\right)\)

Answer 8

hmm ohh... lemme continue that

Answer 9

well... actually it just a simplified form anyhow \(\bf \left(\cfrac{\sqrt{2}}{2}\cdot \cfrac{1}{2}\right)-\left(\cfrac{\sqrt{2}}{2}\cdot\cfrac{\sqrt{3}}{2}\right)\implies
\left(\cfrac{\sqrt{2}}{4}\right)-\left(\cfrac{\sqrt{2\cdot 3}}{4}\right)
\\ \quad \\
\left(\cfrac{\sqrt{2}}{4}\right)-\left(\cfrac{\sqrt{6}}{4}\right)\implies \cfrac{1}{4}\left(\sqrt{2}\right)-\cfrac{1}{4}\left(\sqrt{6}\right)\implies \cfrac{1}{4}(\sqrt{2}-\sqrt{6})
\\ \quad \\
\cfrac{(\sqrt{2}-\sqrt{6})}{4}\)

Answer 10

Oh ok i get it thank you

Answer 11

yw

Answer 12

wrong

Answer 13

the answer is muah and shup shup

Answer 14

Oh thanks max that's helpful lol