Question

Trig....By using the formula for the cosine of the difference of two angles, cos 15 = (√6+√2)/4 and cos 15 degrees= (√2+(√3))/2 (first square root is over whole numerator). Prove that these two quantities are equivalent; i.e., prove (√2+(√3))/2 (first square root is over whole numerator) = (√6+√2)/4. Just like proving identities, you must work with one side only. (Note... it is not acceptable to square a side; i.e., -3 ≠ 3 but 9 = 9.) Trig....By using the formula for the cosine of the difference of two angles, cos 15 = (√6+√2)/4 and cos 15 degrees= (√2+(√3))/2 (first square root is over whole numerator). Prove that these two quantities are equivalent; i.e., prove (√2+(√3))/2 (first square root is over whole numerator) = (√6+√2)/4. Just like proving identities, you must work with one side only. (Note... it is not acceptable to square a side; i.e., -3 ≠ 3 but 9 = 9.) @Mathematics

Answer 1

\[\sqrt{\frac{\sqrt{2}+\sqrt{3}}{2}}=\frac{\sqrt{6}+\sqrt{2}}{4}\]

Answer 2

yep thats the equation we need to prove is equivalent

Answer 3

no there is something wrong here. this is not correct

Answer 4

maybe it should be
\[\frac{\sqrt{\sqrt{2}+\sqrt{3}}}{2}\] on the left hand side

Answer 5

oh my gosh yeah thats it

Answer 6

ok they are still not the same numbers

Answer 7

this is what it is supposed to be
\[\frac{\sqrt{2+\sqrt{3}}}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}\]

Answer 8

and this we can do, although i am sure you are probably gone
\[\frac{\sqrt{2+\sqrt{3}}}{2}\times \frac{2}{2}\]
\[\frac{2\sqrt{2+\sqrt{3}}}{4}\]
\[\frac{\sqrt{4(2+\sqrt{3})}}{4}\]
\[\frac{\sqrt{8+4\sqrt{3}}}{4}\]
\[\frac{\sqrt{6+2\sqrt{12}+2}}{4}\]
\[\frac{\sqrt{(\sqrt{6}+\sqrt{2})^2}}{4}\]
\[\frac{\sqrt{6}+\sqrt{2}}{4}\]