Question

help please!!!!!!!!!!!!!!! 2 and 3, am i on the right track? trig

Answer 1

im supposed to prove 2 and solve 3

Answer 2

Sorry idk :(

Answer 3

@SolomonZelman can you help?

Answer 4

2.\[=\sin ^2\alpha \left( \frac{ 1 }{ \cos \alpha+1 }+\frac{ 1 }{ -\cos \alpha+1 } \right)\]
\[=\sin ^2\alpha \left( \frac{ -\cos \alpha +1+\cos \alpha+1 }{ \left( \cos \alpha +1 \right) \left( -\cos \alpha +1 \right)} \right) =\frac{ 2\sin ^2\alpha }{ 1-\cos ^2 \alpha } =?\]

Answer 5

write again the statement of third.

Answer 6

sqt(1+cos(x))-sqt(2)cos(x/2)=sqt(2)

Answer 7

\[\sqrt{1+\cos x}-\sqrt{2}\cos \frac{ x }{ 2 }=\sqrt{2}\]
i think there is something wrong in your statement.as
\[\sqrt{1+\cos x}-\sqrt{2 \cos ^2 \frac{ x }{ 2 }}=\sqrt{2}\]
\[\sqrt{1+\cos x}-\sqrt{1+\cos x}=\sqrt{2}\]
\[0=\sqrt{2},which~ is~ incorrect.\]

Answer 8

is \[\sqrt{2}\cos(x/2) = \sqrt{2\cos(x/2)}\] @surjithayer

Answer 9

cause the first is how it is written

Answer 10

could you also explain how the second step turns to the third @surjithayer

Answer 11

i dont see how 2cos^2(x/x) turns in to 1+cosx

Answer 12

\[\sqrt{1+\cos x}-\sqrt{2\cos \frac{ x }{ 2 }}=\sqrt{2}\]
\[\sqrt{2\cos ^2\frac{ x }{ 2 }}-\sqrt{2\cos \frac{ x }{ 2 }}=\sqrt{2}\]
\[\sqrt{2}\cos \frac{ x }{ 2 }-\sqrt{2}\sqrt{\cos \frac{ x }{ 2 }}=\sqrt{2}\]
\[\cos \frac{ x }{ 2 }-\sqrt{\cos \frac{ x }{ 2 }}=1\]
\[-\sqrt{\cos \frac{ x }{ 2 }}=1-\cos \frac{ x }{ 2 }\]
squaring both sides
\[\cos \frac{ x }{ 2 }=1+\cos ^2\frac{ x }{ 2 }-2\cos \frac{ x }{ 2 }\]
\[\cos ^2\frac{ x }{ 2}-3 \cos \frac{ x }{ 2 }+1=0\]
\[\cos \frac{ x }{ 2 }=\frac{ 3\pm \sqrt{9-4*1*1} }{ 2*1 }=\frac{ 3\pm \sqrt{5} }{ 2 }\]
\[\left| \cos \frac{ x }{ 2 } \right|\le 1~so~\cos \frac{ x }{ 2 }=\frac{ 3-\sqrt{5} }{ 2 }\]
\[\cos ^2\frac{ x }{ 2 }=\frac{ 9+5-6\sqrt{5} }{ 4 }=\frac{ 7-3\sqrt{5} }{ 2 }\]
\[2 \cos ^2\frac{ x }{ 2 }=7-3\sqrt{5}\]
\[1+\cos x=7-3\sqrt{5},\cos x=6-3\sqrt{5}\]
x=?

Answer 13

\[\cos 2x=\cos \left( x+x \right)=\cos x \cos x-\sin x \sin x=\cos ^2x-\sin ^2x\]
\[=\cos ^2x-\left( 1-\cos ^2x \right)=\cos ^2x-1+\cos ^2x=2\cos ^2x-1\]
so \[\cos 2x=2 \cos ^2x-1\]
put 2x=y,x=y/2
\[\cos y=2\cos ^2\frac{ y }{ 2 }-1,2 \cos ^2\frac{ y }{ 2 }=1+\cos y\]
replace y by x
\[2 \cos ^2\frac{ x }{ 2 }=1+\cos x\]