Half angles bby :)
hola
i did #9 and got 1.
#10 is the one im stuck on, after getting 315 on the angle
\[\cos2\Theta=1/2(1+\cos(2\Theta))\]
\(\Huge\color{blue}{ \sf tan( \frac{225}{2})=\sqrt{ \frac{1-cos(225)}{1+cos(225)} } }\)
\(\Huge\color{blue}{ \sf tan( \frac{225}{2})=\sqrt{ \frac{1-cos(225)}{1+cos(225)} } }\) \(\Huge\color{blue}{ \sf tan( 225)= \frac{sin(225)}{cos(225)}= \frac{sin(180+45)}{cos(180+45)} }\) tell me what you get for tan 225.
sqroot2 / 2
oh crap
-(sqroot2/2 ) / -sqroot2 /2
i got 1
ok, first tell me what does sin 225 equal to, using sin(A+B)=sin A cos B + cos A sin B AND THEN tell me what does cos 225 equal, using cos(A+B)=cos A cos B - sin A sin B
sin a+b = sin 180+45
cos 180 =1 sin 180 = 0 sin 45= cos 45 = √2/2 sin(180+ 45)=sin 180 cos 45 + cos 180 sin 45 =(-1) * √2 / 2 = -√2 / 2 cos(180+45)=cos 180 cos 45 - sin 180 sin 45 = (-1) * √2 / 2 = -√2 / 2 tan (225) = (-√2 / 2) / (-√2 / 2) = 1
yes, it's 1.
lol thanks
yw, still need help with #9 ?
i should be okay. ima start work on 10 now
I got disconnected !
lol its fine
\(\LARGE\color{blue}{ \sf cos(157.5)= \frac{315}{2}= \sqrt{ \frac{1+cos(315)}{2} } }\)
why is there a sqroot over everything if the formula is cos2Θ=1/2(1+cos(2Θ))
nvm i know why
sqroot = 1/2 of whats inside
\(\LARGE\color{blue}{ \sf cos(315)= cos(360-45) }\) \(\Large\color{blue}{ \sf cos(360-45) =cos(360)cos45-sin(360) sin(45)}\) sin 360 = 0 cos 360 = 1 \(\Large\color{blue}{ \sf cos(360-45) =1(√2~/~2)-0 (√2~/~2)}\) \(\Large\color{blue}{ \sf cos(360-45) =√2~/~2}\)
welll its + in the rule but -, but the last part is zero, so doesn't matter.
cosacosb+ (not -) ... EITHER WAY..
whoppsies
you explain so thoroughly trying to find the excact value when i can just glance over at the unit circle and go, "oh look cos315 = sqroot2/2 xD
Thankss
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