Question

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Answer 1

i knew it was a trap :/

Answer 2

haha please help

Answer 3

\[5\sin(2\alpha)=10\sin(\alpha)\cos(\alpha)\]may be a good start

Answer 4

sin(2a) = 2sin(a)cos(a)

and i might think about dividing everything by a sin^2

Answer 5

also note that
\[\cos^2(\alpha)-\sin^2(\alpha)=\cos(2\alpha)\] might actually be a better first step

Answer 6

or that lol

Answer 7

so right away you will get
\[\frac{5\sin(2\alpha)}{\cos(2\alpha)}\] and maybe that will finish

Answer 8

why is 1 and 4 the same?

Answer 9

but as you can see from both our posts, there is no magic bullet. you just have to grind it til you find it, using whatever gimmicks you can
that is why it is called 'trickonometry"

Answer 10

1 and 5 that is

Answer 11

answer is 6. 5 tan x

Answer 12

i see what i did; i used my sin^2 bullet twice. bad amistre!!

Answer 13

no it is not
\[5\tan(\alpha)\] it is
\[5\tan(2\alpha)\] i believe. another trick

Answer 14

\[\cos ^{2}\alpha -\sin ^{2\alpha}=\cos2\alpha\]
\[5\sin2\alpha/\cos2\alpha=5 \tan2\alpha\]

Answer 15

Omg yes i saw wrongly.

Answer 16

this is complicated

Answer 17

oh ok i understand tiaph better, its simpler