Sine Rule Derivation

Use the formulas you’d use to calculate the area of a triangle. See the magic emerge.

$\frac { 1 }{ 2 } bcsinA=\frac { 1 }{ 2 } acsinB\\ \\ bcsinA=acsinB\\ \\ bsinA=asinB\\ \\ \frac { bsinA }{ b } =\frac { asinB }{ b } \\ \\ \frac { sinA }{ 1 } =\frac { asinB }{ b } \\ \\ \frac { sinA }{ 1 } \cdot \frac { 1 }{ a } =\frac { asinB }{ b } \cdot \frac { 1 }{ a } \\ \\ \frac { sinA }{ a } =\frac { sinB }{ b } \\ \\ \\ OR:\\ \\ bsinA=asinB\\ \\ \frac { bsinA }{ sinB } =\frac { asinB }{ sinB } \\ \\ \frac { bsinA }{ sinB } =\frac { a }{ 1 } \\ \\ \frac { bsinA }{ sinB } \cdot \frac { 1 }{ sinA } =\frac { a }{ 1 } \cdot \frac { 1 }{ sinA } \\ \\ \frac { b }{ sinB } =\frac { a }{ sinA }$

Trigonometric Proofs

Rules:

$sec\theta =\frac { 1 }{ cos\theta } \quad \therefore \quad { sec }^{ 2 }\theta =\frac { 1 }{ { cos }^{ 2 }\theta } \\ \\ cosec\theta =\frac { 1 }{ sin\theta } \quad \therefore \quad { cosec }^{ 2 }\theta =\frac { 1 }{ { sin }^{ 2 }\theta } \\ \\ cot\theta =\frac { 1 }{ tan\theta } =\frac { 1 }{ \frac { sin\theta }{ cos\theta } } =\frac { cos\theta }{ sin\theta } \\ \\ \therefore \quad { cot }^{ 2 }\theta =\frac { 1 }{ { tan }^{ 2 }\theta } =\frac { { cos }^{ 2 }\theta }{ { sin }^{ 2 }\theta }$

Transformation 1:

${ sin }^{ 2 }\theta +{ cos }^{ 2 }\theta =1\\ \\ \frac { { sin }^{ 2 }\theta }{ { sin }^{ 2 }\theta } +\frac { { cos }^{ 2 }\theta }{ { sin }^{ 2 }\theta } =\frac { 1 }{ { sin }^{ 2 }\theta } \\ \\ 1+{ cot }^{ 2 }\theta ={ cosec }^{ 2 }\theta$

Transformation 2:

${ sin }^{ 2 }\theta +{ cos }^{ 2 }\theta =1\\ \\ \frac { { sin }^{ 2 }\theta }{ { cos }^{ 2 }\theta } +\frac { { cos }^{ 2 }\theta }{ { cos }^{ 2 }\theta } =\frac { 1 }{ { cos }^{ 2 }\theta } \\ \\ { tan }^{ 2 }\theta +1={ sec }^{ 2 }\theta$

Derivative of y=cotx

$y=cotx=\frac { cosx }{ sinx } =\frac { u }{ v } ,\\ \\ If\quad y=\frac { u }{ v } ,\quad \frac { dy }{ dx } =\frac { v\frac { du }{ dx } -u\frac { dv }{ dx } }{ { v }^{ 2 } } .\\ \\ u=cosx,\quad \therefore \quad \frac { du }{ dx } =-sinx\\ v=sinx,\quad \therefore \quad \frac { dv }{ dx } =cosx,\quad { v }^{ 2 }=\sin ^{ 2 }{ x } \\ \\ So:\\ \\ \frac { dy }{ dx } =\frac { sinx\cdot \left( -sinx \right) -cosxcosx }{ \sin ^{ 2 }{ x } } \\ \\ =\frac { -\sin ^{ 2 }{ x } -\cos ^{ 2 }{ x } }{ \sin ^{ 2 }{ x } } =\frac { -\sin ^{ 2 }{ x } }{ \sin ^{ 2 }{ x } } -\frac { \cos ^{ 2 }{ x } }{ \sin ^{ 2 }{ x } } \\ \\ =-1-\cot ^{ 2 }{ x } =-{ cosec }^{ 2 }x\\ \\ \therefore \quad If\quad y=cotx,\quad \frac { dy }{ dx } =-{ cosec }^{ 2 }x\\ \\ \\$

tan(A-B)=(tanA-tanB)/(1+tanAtanB)

Prove that:

$tan\left( A-B \right) =\frac { tanA-tanB }{ 1+tanAtanB }$

Firstly:

$tan\left( \alpha +\beta \right) =\frac { tan\alpha +tan\beta }{ 1-tan\alpha tan\beta }$

Also, remember that:

$tan\left( -\theta \right) =-tan\theta$

So:

$\beta =-\phi \\ \\ tan\left( \alpha +\left( -\phi \right) \right) =\frac { tan\alpha +tan\left( -\phi \right) }{ 1-tan\alpha tan\left( -\phi \right) } \\ \\ =\frac { tan\alpha -tan\phi }{ 1+tan\alpha tan\phi } =tan\left( \alpha -\phi \right) \\ \\ \therefore \quad tan\left( A-B \right) =\frac { tanA-tanB }{ 1+tanAtanB }$

Derivate Of y=cosx Proof

Prove that:

$If\quad f(x)=cosx,\quad f$

$\lim _{ \delta x\rightarrow 0 }{ \frac { f\left( x+\delta x \right) -f\left( x \right) }{ \delta x } } \\ \\ =\lim _{ \delta x\rightarrow 0 }{ \frac { cos\left( x+\delta x \right) -cosx }{ \delta x } } \\ \\ =\lim _{ \delta x\rightarrow 0 }{ \frac { cosxcos\delta x-sinxsin\delta x-cosx }{ \delta x } } \\ \\ =\lim _{ \delta x\rightarrow 0 }{ \frac { cosx\left( cos\delta x-1 \right) -sinxsin\delta x }{ \delta x } } \\ \\ =\lim _{ \delta x\rightarrow 0 }{ \frac { cosx\left( cos\delta x-1 \right) }{ \delta x } } -\frac { sinxsin\delta x }{ \delta x } \\ \\ =0-sinx=-sinx\\ \\ As:\\ \\ \lim _{ \delta x\rightarrow 0 }{ \frac { \left( cos\delta x-1 \right) }{ \delta x } } =0\\ \\ \lim _{ \delta x\rightarrow 0 }{ \frac { sin\delta x }{ \delta x } } =1$

tan(A+B)=(tanA+tanB)/(1-tanAtanB)

Prove that:

$tan(A+B)=\frac { tanA+tanB }{ 1-tanAtanB }$

$LHS=tan(A+B)=\frac { sin(A+B) }{ cos(A+B) } \\ \\ =\frac { sinAcosB+cosAsinB }{ cosAcosB-sinAsinB } =\frac { \frac { sinA }{ 1 } \cdot \frac { sinB }{ tanB } +\frac { sinA }{ tanA } \cdot \frac { sinB }{ 1 } }{ \frac { sinA }{ tanA } \cdot \frac { sinB }{ tanB } -\frac { sinAsinB }{ 1 } } \\ \\ =\frac { \frac { sinAsinB }{ tanB } +\frac { sinAsinB }{ tanA } }{ \frac { sinAsinB }{ tanAtanB } -\frac { sinAsinB }{ 1 } } =\frac { sinAsinB\left( \frac { 1 }{ tanB } +\frac { 1 }{ tanA } \right) }{ sinAsinB\left( \frac { 1 }{ tanAtanB } -\frac { 1 }{ 1 } \right) } \\ \\ =\frac { \frac { 1 }{ tanB } +\frac { 1 }{ tanA } }{ \frac { 1 }{ tanAtanB } -\frac { 1 }{ 1 } } =\frac { \frac { tanA+tanB }{ tanAtanB } }{ \frac { 1-tanAtanB }{ tanAtanB } } =\frac { tanA+tanB }{ tanAtanB } \cdot \frac { tanAtanB }{ 1-tanAtanB } \\ \\ =\frac { tanA+tanB }{ 1-tanAtanB } =RHS$

ArcSin, ArcCos & ArcTan

Learn how to produce the arcsin, arccos and arctan graphs using the unit circle.