Question

if R vector = A vector + B vector show that R^2 = A^2+B^2+2AB cos theta​

Answer 1

According to sum of vectors : $\vec{R},\vec{A}.\vec{B}\:such\:that\:\vec{R}=\vec{A}+\vec{B}$

This means that this has to be a triangle with ABR .

We know that the cosine rule states that :

R² = A² + B² - 2 AB  cos ( 180° - θ )

⇒ R² = A² + B² - 2 AB cos ( 90° - ( - 90 + θ ) )

⇒ R² = A² + B² - 2 AB ( - sin ( 90 - θ ) )

⇒ R² = A² + B² - 2 AB ( - cos θ )

⇒ R² = A² + B² + 2 AB cos θ

Hence proved !

Explanation :

cos θ is negative in the 2 nd quadrant .

The formula used in trigonometry are :

cos ( 90 - θ ) = sin θ

sin ( 90 - θ ) = cos θ

Cosine rule is :

R² = A² + B² + 2 AB cos θ where θ is the opposite angle of R in a triangle .

Answer 2

ANSWER:--------------

{ R² = A² + B² - 2 }

[AB  cos { 180° - θ )

===>note here ,the follows

===> R² = A² + B² - 2 AB cos ( 90 - !(- 90 + θ )

cøs(9∅!!!!(-90+(∅)

===> R² = A² + B² - 2 AB - sin { 90 - θ }

===> R² = A² + B² - 2 AB - cos θ

===>R² = A² + B² + 2 AB cos θ

proved this problem:-&

hope it helps:--

T!—!ANKS!!!!