Proving Double-Angle Formulae for Triangles | Helpful Proof Guide

[odolwa99]

In the final proof, I have \frac{h-h\tan B}{\tan B}, rather than \frac{h-h\tan B}{1+\tan B}. Can anyone help me out?

Many thanks.

Homework Statement

In the triangle pqr (see attachment), |\angle qrp|=90^{\circ} & |rp| = h. s is a point on [qr] such that |\angle spq|=2B & |\angle rps|=45^{\circ}-B, 0^{\circ}<B<45^{\circ}. Show that |sr|=h \tan(45^{\circ}-B)

Homework Equations

The Attempt at a Solution

h\tan(45-B)=h(\frac{\tan45-\tan B}{1+ tan45\cdot\tan B})=\frac{h-h\tan B}{1+\tan B}

Apply the sine rule: \frac{\sin(45-B)}{|sr|}=\frac{\sin(180-(90+45-B))}{h}
h(\sin(45-B))=|sr|\sin(45+B)
h(\sin\cos B-\cos45\sin B)=|sr|(\sin45\cos B+\cos45\sin B)
h(\frac{\cos B}{\sqrt{2}}-frac{sin B}{\sqrt{2}})=|sr|(\frac{\cos B}{\sqrt{2}}+\frac{\sin B}{\sqrt{2}})
|sr|=(\frac{h\cos B-h\sin B}{\sqrt{2}})/(\frac{\cos B}{\sin B}{\sqrt{2}})
\frac{h\sqrt{2}(\cos B- \sin B)}{\sqrt{2}(\cos B+\sin B})
h(\frac{\sin B}{\cos B}+\frac{\cos B}{\sin B})
h(-\tan B+\frac{1}{\tan B})
\frac{h-h\tan B}{\tan B}

 

[LCKurtz]

What am I missing? I see right triangle srp with$$
\tan(45-B) = \frac {sr} h$$Just solve for $sr$.

 

[odolwa99]

Looks like I was over-thinking this. Ok, thank you.

 

[SammyS]

In the final proof, I have \frac{h-h\tan B}{\tan B}, rather than \frac{h-h\tan B}{1+\tan B}. Can anyone help me out?

Many thanks.

Homework Statement

In the triangle pqr (see attachment), |\angle qrp|=90^{\circ} & |rp| = h. s is a point on [qr] such that |\angle spq|=2B & |\angle rps|=45^{\circ}-B, 0^{\circ}<B<45^{\circ}. Show that |sr|=h \tan(45^{\circ}-B)

Homework Equations

The Attempt at a Solution

h\tan(45-B)=h(\frac{\tan45-\tan B}{1+ tan45\cdot\tan B})=\frac{h-h\tan B}{1+\tan B}

Apply the sine rule: \frac{\sin(45-B)}{|sr|}=\frac{\sin(180-(90+45-B))}{h}
h(\sin(45-B))=|sr|\sin(45+B)
h(\sin\cos B-\cos45\sin B)=|sr|(\sin45\cos B+\cos45\sin B)
h(\frac{\cos B}{\sqrt{2}}-\frac{\sin B}{\sqrt{2}})=|sr|(\frac{\cos B}{\sqrt{2}}+\frac{\sin B}{\sqrt{2}})
|sr|=(\frac{h\cos B-h\sin B}{\sqrt{2}})/(\frac{\cos B}{\sin B}{\sqrt{2}})
\frac{h\sqrt{2}(\cos B- \sin B)}{\sqrt{2}(\cos B+\sin B})
h(\frac{\sin B}{\cos B}+\frac{\cos B}{\sin B})
h(-\tan B+\frac{1}{\tan B})
\frac{h-h\tan B}{\tan B}

\displaystyle \frac{\cos B- \sin B}{\cos B+\sin B}\ne\frac{-\sin B}{\cos B}+\frac{\cos B}{\sin B}

You can't break up a fraction in that manner!