Question

Find an exact value.

tan 15°

I really appreciate the help! THANK YOU :)

Answer 1

use the difference of 2 angles

\[\tan(45 - 30) = \frac{\tan(45) - \tan(30)}{1 + \tan(45)\tan(30)}\]

now just substitute the exact values that you know

Answer 2

I got this wrong... I guess the answer isn't 2 - sqrt3 /4

Answer 3

hi jdoe I appreciate you taking the time to look at my problem :)

Answer 4

$$\bf
tan(45^o - 30^o) = \frac{tan(45) - tan(30)}{1 + tan(45)tan(30)}\\
tan(45^o - 30^o) = \cfrac{1 -\frac{1}{\sqrt{3}} }{1 + (1)\left(\frac{1}{\sqrt{3}}\right)}\\
tan(45^o - 30^o) = \large \cfrac{\frac{\sqrt{3}-1}{\sqrt{3}} }{\frac{\sqrt{3}+1}{\sqrt{3}}}\\
$$

Answer 5

:)

Answer 6

that isn't one of the options though...argh I missed five questions on this assignment and I want to know what I did wrong before I retake it...want me to show you what the options were?

Answer 7

well, you still need to simplify there further

Answer 8

recall that => \(\bf \large \cfrac{\frac{a}{b}}{\frac{c}{d}} \implies \frac{a}{b} \times \frac{d}{c}\)

Answer 9

have you rationalised the denominator

you are looking at

\[\frac{\sqrt{3} -1}{\sqrt{3} + 1} \times \frac{\sqrt{3} -1}{\sqrt{3} -1} = \frac{3 - 2\sqrt{3} + 1}{3 - 1}\]

Answer 10

so the final solution is

\[2 - \sqrt{3}\]

is that an option

Answer 11

yes it is! thank you! simplifying! ah got it!

Answer 12

so as @jdoe0001 said

\[\frac{\frac{\sqrt{3} -1}{\sqrt{3}}}{\frac{\sqrt{3} +1}{\sqrt{3}}} = \frac{\sqrt{3} -1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}+1}\]

then you get what I have above

Answer 13

I'm still trying to see how you got the 2 though? when you simplfied?