Rejecting double angled formula answers, why ?

[leflyi]

Quick question!


Use double angled formulas to show that

Explain why answers need to be rejected.

I proved the fact, just a query on why answers must be rejected?

Maybe i proved incorrectly

My working:

[leflyi]

Nevermind


I transposed tan's double angled formula into a quadractic expression, and expressed how tan < 0, which mean 1+root(2) had to be discounted or rejected

[lzxnl]

May I just add that I have a tan half angle formula that is probably a lot easier to use.
tan x/2=sin x/2 / cos x/2  =  2 sin^2 x/2  /  2 sin x/2 cos x/2 = ( 1-(1-2 sin^2 x/2) ) / sin 2(x/2) = (1-cos x)/sin x
Using x/2=-pi/8, x=-pi/4, we have tan x/2=(1-cos -pi/4)/(sin -pi/4)= (1-1/2*sqrt 2)/(-1/2*sqrt 2)=1-sqrt 2

Unfortunately sine and cosine half angle formulas from the double angle formulas DO require square roots, so you'll have to use your discretion with those.

[Holmes]

Glad to know that you figured it out, although you're not actually meant to use what you're trying to prove in your actual proof. You were solving for tan(-pi/8) so you have to figure that out using the double angle formula, you can't sub in (-) into it like you've done in your proof.



Let



Let





Then you just solve for u which is equal to tan(-pi/8). Your answer is the negative root because tan(-pi/8) is in the 4th quadrant and tan is negative in that quadrant.

[leflyi]

Thats how I prooved above ^^