Question

In ΔABC, For any angle θ, then show that a cos θ = b cos(C + θ) + c cos(B - θ).

Answer 1

Answer:

Step-by-step explanation:

Formula used:

cos(A+B)=cosA cosB - sinA sinB

cos(A-B)=cosA cosB + sinA sinB

Area of triangle ABC,

$\Delta=\frac{1}{2}ab\:sinC=\frac{1}{2}ac\:sinB$

$\frac{\Delta}{a}=b\:sinC=\frac{1}{2}c\:sinB$

$b\:cos(C+\theta)+c\:cos(B-\theta)\\\\= b[cosC\:cos\theta-sinC\:sin\theta]+c[cosB\:cos\theta+sinB\:sin\theta]$

$= b\:cosC\:cos\theta-b\:sinC\:sin\theta+c\:cosB\:cos\theta+c\:sinB\:sin\theta\\\\= (b\:cosC+c\:cosB)cos\theta+(-b\:sinC+c\:sinB)sin\theta$

$= acos\theta+(-\frac{\Delta}{a}+\frac{\Delta}{a})sin\theta\\= acos\theta+(0)sin\theta\\= acos\theta$

Answer 2

Answer:

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