MENU
No registration required!
Trigonometric Prove
5 posts
• Page 1 of 1
Re: Trigonometric Prove
by Guest » Fri Aug 28, 2026 11:46 pm
cos 2[tex]\varphi[/tex]= cos 2[tex]\varphi[/tex]
cos 2[tex]\varphi[/tex]= [tex]cos^{2 } \varphi -sin^{2 } \varphi[/tex]
cos 2[tex]\varphi[/tex]= [tex]\frac{ cos^{2 } \varphi - sin^{2 } \varphi }{1}[/tex]
cos 2[tex]\varphi[/tex]= [tex]\frac{ cos^{2 } \varphi -sin^{2 } \varphi }{ cos^{2 } \varphi + sin^{2 } \varphi } . \frac{ cos^{2 } \varphi }{ cos^{2 } \varphi }[/tex]
cos 2[tex]\varphi[/tex]=[tex]\frac{ \frac{ cos^{2 } \varphi -sin^{2 } \varphi }{ cos^{2 } \varphi } }{ \frac{ cos^{2 } \varphi +sin^{2 } \varphi }{ cos^{2 } \varphi } }[/tex]
cos 2[tex]\varphi[/tex]= [tex]\frac{1 -( \frac{sin \varphi }{cos \varphi }) ^{2 } }{1 + (\frac{sin \varphi }{cos \varphi }) ^{2 } }[/tex]
cos 2[tex]\varphi[/tex]= [tex]\frac{1- tg^{2 } \varphi }{1+ tg^{2 } \varphi }[/tex]
cos 2[tex]\varphi[/tex]= [tex]\frac{1- z^{2 } }{1+ z^{2 } }[/tex]
cos 2[tex]\varphi[/tex]= [tex]cos^{2 } \varphi -sin^{2 } \varphi[/tex]
cos 2[tex]\varphi[/tex]= [tex]\frac{ cos^{2 } \varphi - sin^{2 } \varphi }{1}[/tex]
cos 2[tex]\varphi[/tex]= [tex]\frac{ cos^{2 } \varphi -sin^{2 } \varphi }{ cos^{2 } \varphi + sin^{2 } \varphi } . \frac{ cos^{2 } \varphi }{ cos^{2 } \varphi }[/tex]
cos 2[tex]\varphi[/tex]=[tex]\frac{ \frac{ cos^{2 } \varphi -sin^{2 } \varphi }{ cos^{2 } \varphi } }{ \frac{ cos^{2 } \varphi +sin^{2 } \varphi }{ cos^{2 } \varphi } }[/tex]
cos 2[tex]\varphi[/tex]= [tex]\frac{1 -( \frac{sin \varphi }{cos \varphi }) ^{2 } }{1 + (\frac{sin \varphi }{cos \varphi }) ^{2 } }[/tex]
cos 2[tex]\varphi[/tex]= [tex]\frac{1- tg^{2 } \varphi }{1+ tg^{2 } \varphi }[/tex]
cos 2[tex]\varphi[/tex]= [tex]\frac{1- z^{2 } }{1+ z^{2 } }[/tex]
- Guest
Re: Trigonometric Prove
by nycmath » Sat Aug 29, 2026 4:26 am
Guest wrote:cos 2[tex]\varphi[/tex]= cos 2[tex]\varphi[/tex]
cos 2[tex]\varphi[/tex]= [tex]cos^{2 } \varphi -sin^{2 } \varphi[/tex]
cos 2[tex]\varphi[/tex]= [tex]\frac{ cos^{2 } \varphi - sin^{2 } \varphi }{1}[/tex]
cos 2[tex]\varphi[/tex]= [tex]\frac{ cos^{2 } \varphi -sin^{2 } \varphi }{ cos^{2 } \varphi + sin^{2 } \varphi } . \frac{ cos^{2 } \varphi }{ cos^{2 } \varphi }[/tex]
cos 2[tex]\varphi[/tex]=[tex]\frac{ \frac{ cos^{2 } \varphi -sin^{2 } \varphi }{ cos^{2 } \varphi } }{ \frac{ cos^{2 } \varphi +sin^{2 } \varphi }{ cos^{2 } \varphi } }[/tex]
cos 2[tex]\varphi[/tex]= [tex]\frac{1 -( \frac{sin \varphi }{cos \varphi }) ^{2 } }{1 + (\frac{sin \varphi }{cos \varphi }) ^{2 } }[/tex]
cos 2[tex]\varphi[/tex]= [tex]\frac{1- tg^{2 } \varphi }{1+ tg^{2 } \varphi }[/tex]
cos 2[tex]\varphi[/tex]= [tex]\frac{1- z^{2 } }{1+ z^{2 } }[/tex]
Why don't you just join the site?
Re: Trigonometric Prove
by Eigenvalue » Sat Sep 19, 2026 1:35 am
Let z=tan(x)
1. Rewrite tan(x) as [tex]\frac{sin(x)}{cos(x)}[/tex]
[tex]\frac{sin(x}{cos(x)}[/tex]=z=[tex]\frac{z}{1}[/tex]
hypotenuse: [tex]\sqrt{z²+1}[/tex]
sin(x)=[tex]\frac{z}{ \sqrt{z²+1} }[/tex]
cos(x)=[tex]\frac{1}{ \sqrt{z²+1} }[/tex]
2. Rewrite cos(2x) as cos²(x)-sin²(x)
cos²(x)-sin²(x)
=([tex]\frac{1}{ \sqrt{z²+1} }[/tex])²-([tex]\frac{z}{ \sqrt{z²+1} }[/tex])
=[tex]\frac{1}{z²+1}[/tex]-[tex]\frac{z²}{z²+1}[/tex]
=[tex]\frac{1-z²}{z²+1}[/tex]
3. Rewrite sin(2x)=2sin(x)cos(x)
2sin(x)cos(x)
=2([tex]\frac{z}{ \sqrt{z²+1} }[/tex])([tex]\frac{1}{z²+1)}[/tex]
=2([tex]\frac{z}{z²+1}[/tex])
=[tex]\frac{2z}{z²+1}[/tex]
1. Rewrite tan(x) as [tex]\frac{sin(x)}{cos(x)}[/tex]
[tex]\frac{sin(x}{cos(x)}[/tex]=z=[tex]\frac{z}{1}[/tex]
hypotenuse: [tex]\sqrt{z²+1}[/tex]
sin(x)=[tex]\frac{z}{ \sqrt{z²+1} }[/tex]
cos(x)=[tex]\frac{1}{ \sqrt{z²+1} }[/tex]
2. Rewrite cos(2x) as cos²(x)-sin²(x)
cos²(x)-sin²(x)
=([tex]\frac{1}{ \sqrt{z²+1} }[/tex])²-([tex]\frac{z}{ \sqrt{z²+1} }[/tex])
=[tex]\frac{1}{z²+1}[/tex]-[tex]\frac{z²}{z²+1}[/tex]
=[tex]\frac{1-z²}{z²+1}[/tex]
3. Rewrite sin(2x)=2sin(x)cos(x)
2sin(x)cos(x)
=2([tex]\frac{z}{ \sqrt{z²+1} }[/tex])([tex]\frac{1}{z²+1)}[/tex]
=2([tex]\frac{z}{z²+1}[/tex])
=[tex]\frac{2z}{z²+1}[/tex]
- Eigenvalue
- Posts: 1041
- Joined: Mon Aug 10, 2026 10:38 pm
- Reputation: 382
Re: Trigonometric Prove
by nycmath » Sat Sep 19, 2026 9:09 am
Eigenvalue wrote:Let z=tan(x)
1. Rewrite tan(x) as [tex]\frac{sin(x)}{cos(x)}[/tex]
[tex]\frac{sin(x}{cos(x)}[/tex]=z=[tex]\frac{z}{1}[/tex]
hypotenuse: [tex]\sqrt{z²+1}[/tex]
sin(x)=[tex]\frac{z}{ \sqrt{z²+1} }[/tex]
cos(x)=[tex]\frac{1}{ \sqrt{z²+1} }[/tex]
2. Rewrite cos(2x) as cos²(x)-sin²(x)
cos²(x)-sin²(x)
=([tex]\frac{1}{ \sqrt{z²+1} }[/tex])²-([tex]\frac{z}{ \sqrt{z²+1} }[/tex])
=[tex]\frac{1}{z²+1}[/tex]-[tex]\frac{z²}{z²+1}[/tex]
=[tex]\frac{1-z²}{z²+1}[/tex]
3. Rewrite sin(2x)=2sin(x)cos(x)
2sin(x)cos(x)
=2([tex]\frac{z}{ \sqrt{z²+1} }[/tex])([tex]\frac{1}{z²+1)}[/tex]
=2([tex]\frac{z}{z²+1}[/tex])
=[tex]\frac{2z}{z²+1}[/tex]
Terrific! A job well-done!
5 posts
• Page 1 of 1
Return to Trigonometry, Pythagoras' Theorem, Sine Rule, Cosine Rule
Who is online
Users browsing this forum: No registered users and 1 guest