you know two sides of the triangle..
lets work on that cosine
(2+sqrt2)^2=4+2sqrt2+2=6+2sqrt2
the third side of triangle..
sqrt((2+sqrt2)^2 +2^2)
sqrt(10+2sqrt2)
check that. If correct, then
sine 67.5=(10+2sqrt2)/2
Given cos 67.5° = [√(2+√2)]/2, find tan 67.5° , simplify where needed, and show work.
I'm starting to learn this stuff, and I'm so confused where to start. I know they gave me the coordinate X as in cos 67.5° = [√(2+√2)]/2, and I also know that tan = sin/cos, but in order for me to get there, I'll need to know sin first right? Please help me.
2 answers
First of all, you have a typo:
cos 67.5 = √(2-√2) /2
Have you come across the identity
(sinØ)^2 + (cosØ)^2 = 1 ?
We can apply it here:
(sin67.5)^2 + (2 - √2)/4 = 1
sin67.5 ^2 = 1 - (2 - √2)/4
= ( 4 - 2 + √2)/4
= (2 + √2)/4
sin67.5 = [√(2+√2)]/2
so tan67.5
= sin67.5/cos67.5
= [√(2+√2)]/2 / [√(2-√2)]/2
= √(2+√2) / √(2-√2)
or
= √[ (2+√2)/(2-√2) ]
take over
cos 67.5 = √(2-√2) /2
Have you come across the identity
(sinØ)^2 + (cosØ)^2 = 1 ?
We can apply it here:
(sin67.5)^2 + (2 - √2)/4 = 1
sin67.5 ^2 = 1 - (2 - √2)/4
= ( 4 - 2 + √2)/4
= (2 + √2)/4
sin67.5 = [√(2+√2)]/2
so tan67.5
= sin67.5/cos67.5
= [√(2+√2)]/2 / [√(2-√2)]/2
= √(2+√2) / √(2-√2)
or
= √[ (2+√2)/(2-√2) ]
take over
1 answer